CUET UG Mathematics Booster Test 1 - Applications of Determinants in Coordinate Geometry -

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CUET UG Mathematics Booster Test 1 - Applications of Determinants in Coordinate Geometry

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Mathematics Booster Test 1 - Applications of Determinants in Coordinate Geometry

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QUESTION 1 OF 20

Which of the following correctly express the area of a triangle with vertices \(\left(x_{1},\ y_{1}\right)\), \(\left(x_{2},\ y_{2}\right)\), \(\left(x_{3},\ y_{3}\right)\)?
I. \(\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]\)
II. \(\frac{1}{2}∣∣\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1\end{pmatrix}∣∣\)
III. \(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}\)

Answer Review

1 Which of the following correctly express the area of a triangle with vertices \(\left(x_{1},\ y_{1}\right)\), \(\left(x_{2},\ y_{2}\right)\), \(\left(x_{3},\ y_{3}\right)\)?
I. \(\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]\)
II. \(\frac{1}{2}∣∣\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1\end{pmatrix}∣∣\)
III. \(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}\)

A I only

B I and II only

C II and III only

D I, II and III

💡 Short Explanation

Coordinate area formula expands determinant Determinant form and expanded form are equivalent Expression III is incomplete

✅ Correct Answer Explanation

Statement I is the expanded coordinate formula for triangle area. Statement II is the determinant representation of the same formula. Statement III omits subtraction terms: \(-(x_{1}y_{3}+x_{2}y_{1}+x_{3}y_{2})\) so it is incomplete. Therefore only Statements I and II are correct, making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Ignores the determinant form, which is also a standard area formula.
Option C → Statement III is incomplete and cannot independently represent area.
Option D → All three cannot be correct because Statement III lacks necessary negative terms.

🎯 Strategy

Used

Option Grouping

Application:

Check whether each statement matches the standard determinant area formula.

Final Logic:

Only Statements I and II correctly represent triangle area.

🧠 Memory Trick

"Determinant and coordinate forms are twins"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Area of Triangle

2 Match the determinant condition with geometric meaning:

List I	List II
1. Determinant = 0	a. Absolute area of the triangle
2. Determinant > 0	b. Signed area (clockwise orientation)
3. Determinant < 0	c. Signed area (counter-clockwise orientation)
4. \(∣Determinant∣\)	d. Points are collinear

A 1–a, 2–b, 3–c, 4–d

B 1–b, 2–a, 3–d, 4–c

C 1–d, 2–c, 3–b, 4–a

D 1–c, 2–d, 3–a, 4–b

💡 Short Explanation

Zero determinant implies collinearity Sign indicates orientation Absolute value gives geometric area

✅ Correct Answer Explanation

A zero determinant means area is zero, so points are collinear. Positive determinant indicates counter-clockwise orientation, while negative determinant represents clockwise orientation. Absolute value removes sign dependence and gives geometric area. Therefore: \(1-d, 2-c, 3-b, 4-a\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Collinearity cannot correspond to absolute area.
Option B → Positive determinant does not directly represent absolute area.
Option D → Zero determinant never represents counter-clockwise orientation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match determinant sign properties with geometric interpretations.

Final Logic:

Determinant sign determines orientation while modulus gives area.

🧠 Memory Trick

"Zero → line, modulus → area"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Geometric Interpretation

3 Identify the incorrect statement:

A Area is always positive

B Determinant value can be negative

C Absolute value is applied to determinant

D Absolute value is needed only when coordinates are negative

💡 Short Explanation

Area is always non-negative Determinants may be positive or negative Absolute value depends on orientation, not coordinates

✅ Correct Answer Explanation

Absolute value is applied because determinant sign depends on orientation of points, not whether coordinates are negative. Therefore Option D is incorrect. Options A, B, and C correctly describe determinant-area properties. Geometric area must always be non-negative.

❓ Why Other Options Are Incorrect

Option A → Correct because geometric area cannot be negative.
Option B → Correct since determinant sign changes with orientation.
Option C → Correct because area uses modulus of determinant value.

🎯 Strategy

Used

Extreme Word Filter

Application:

The word "only" in Option D makes the statement too restrictive.

Final Logic:

Absolute value depends on orientation, not coordinate sign alone.

🧠 Memory Trick

"Orientation changes sign, modulus fixes area"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Absolute Area

4 The area of a triangle with vertices \(\left(k,\ 0\right)\), \(\left(2,\ 0\right)\), and \(\left(0,\ 5\right)\)is 5 sq units.
Find the positive difference between possible values of \(k\).

A 2

B 4

C 6

D 8

💡 Short Explanation

Use determinant area formula Solve absolute value equation Difference between roots equals 4

✅ Correct Answer Explanation

Area: \(\frac{1}{2}∣5(k-2)∣=5∣k-2∣=2\) Thus: \(k=4 or 0\) Positive difference: \(4-0=4\) Therefore the correct answer is B) 4. The provided answer A) 2 is incorrect.

❓ Why Other Options Are Incorrect

Option A → Gives incorrect difference between obtained values.
Option C → No valid area equation produces difference 6.
Option D → Exceeds actual separation between possible values.

🎯 Strategy

Used

Substitution

Application:

Apply area formula and solve resulting modulus equation.

Final Logic:

Possible values are 0 and 4, whose difference is 4.

🧠 Memory Trick

"Area equation → modulus roots"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Area of Triangle

5 If three points lie on the x-axis, the area of the triangle formed is:

A 1 sq unit

B Infinite

C 0 sq units

D Cannot be determined

💡 Short Explanation

Points on same line are collinear Collinear points form zero area Determinant becomes zero

✅ Correct Answer Explanation

Points on the x-axis have the same y-coordinate, so they are collinear. A triangle formed by collinear points has zero area. Hence Option C is correct. Determinant evaluation also gives zero in such cases.

❓ Why Other Options Are Incorrect

Option A → Collinear points cannot enclose positive area.
Option B → Triangle area cannot become infinite.
Option D → Area is completely determined from collinearity.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use geometric interpretation of collinear points.

Final Logic:

Same-line points always produce zero area.

🧠 Memory Trick

"Same line → zero triangle"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Collinearity

6 If three points lie on the line \(y=2x+1\), the determinant formed by them is:

A 3

B 1

C −1

D 0

💡 Short Explanation

All points lie on same straight line Collinear points give zero area Determinant becomes zero

✅ Correct Answer Explanation

Any three points lying on the same line are collinear. The determinant used for triangle area becomes zero because the enclosed area is zero. Hence Option D is correct. Nonzero determinants would indicate a genuine triangle.

❓ Why Other Options Are Incorrect

Option A → Positive determinant implies non-collinear points.
Option B → Area cannot remain nonzero for collinear points.
Option C → Negative determinant also indicates nonzero signed area.

🎯 Strategy

Used

Elimination

Application:

Use the property that collinear points always give zero determinant.

Final Logic:

Points on one line imply determinant equals zero.

🧠 Memory Trick

"Collinear ⇒ determinant zero"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Collinearity Test

7 Points are \(\left(0,\ 0\right)\), \(\left(2,\ 0\right)\), and \(\left(k,\ 2\right)\), where \(k\in \{0,1,2,3\}\).
Find the probability that the area of the triangle is greater than 2 sq units.

A 0.5

B 0.25

C 0.75

D 0.0

💡 Short Explanation

Base remains constant Height equals 2 always Area never exceeds 2

✅ Correct Answer Explanation

Using: \(Area=\frac{1}{2}\times 2\times 2=2\) For all values of \(k\), the base between \(\left(0,\ 0\right)\)and \(\left(2,\ 0\right)\)is fixed at 2, and height from \(\left(k,\ 2\right)\)is always 2. Thus area always equals 2, never greater than 2. Therefore probability is 0.

❓ Why Other Options Are Incorrect

Option A → Area is never greater than 2 for any allowed value of \(k\).
Option B → No single value satisfies the required condition.
Option C → Three successful outcomes do not exist.

🎯 Strategy

Used

Substitution

Application:

Compute triangle area for the general point \(\left(k,\ 2\right)\).

Final Logic:

Area remains constant at 2 for every allowed value.

🧠 Memory Trick

"Fixed base + fixed height = fixed area"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Applications of Area

8 If a parallelogram has area 10, then the triangle formed using the same vectors has area:

A 10

B 5

C 20

D 0

💡 Short Explanation

Triangle shares same base and height Triangle area is half parallelogram area Divide by 2

✅ Correct Answer Explanation

Area of a parallelogram formed by two vectors equals: \(∣\vec{a}\times \vec{b}∣\) Area of the corresponding triangle is half: \(\frac{1}{2}∣\vec{a}\times \vec{b}∣\) Thus: \(\frac{1}{2}(10)=5\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Triangle area is not equal to parallelogram area.
Option C → Triangle area cannot exceed parallelogram area formed by same vectors.
Option D → Nonzero vectors produce nonzero area.

🎯 Strategy

Used

Dimensional/Unit Analysis

Application:

Use the geometric relation between parallelogram and triangle areas.

Final Logic:

Triangle area is exactly half of parallelogram area.

🧠 Memory Trick

"Triangle = half parallelogram"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Geometric Applications

9 If the determinant value obtained is -8, then the geometric area is:

A -8

B 8

C 4

D 16

💡 Short Explanation

Geometric area uses modulus Negative sign shows orientation only Absolute value of -8 is 8

✅ Correct Answer Explanation

Geometric area is always non-negative. Therefore determinant value must be taken in absolute value: \(∣-8∣=8\) Hence Option B is correct. The provided answer C) 4 is incorrect because no division by 2 was specified in the question.

❓ Why Other Options Are Incorrect

Option A → Area cannot remain negative.
Option C → Division by 2 is not justified from the given statement.
Option D → Area is not doubled during modulus operation.

🎯 Strategy

Used

Elimination

Application:

Remove impossible negative-area choices and apply modulus rule.

Final Logic:

Absolute value converts \(-8\) into geometric area \(8\).

🧠 Memory Trick

"Geometry ignores determinant sign"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Absolute Area

10 The area under the line joining \(\left(0,\ 0\right)\)and \(\left(2,\ 4\right)\)from \(x=0\) to \(x=2\) is equal to the determinant area of triangle with vertices \(\left(0,\ 0\right)\), \(\left(2,\ 0\right)\), and \(\left(2,\ 4\right)\).
Find the area:

A 2

B 4

C 6

D 8

💡 Short Explanation

Triangle base equals 2 Height equals 4 Area formula gives 4

✅ Correct Answer Explanation

The triangle has vertices: \(\left(0,0), (2,0), (2,4\right)\) Base: \(2\) Height: \(4\) Area: \(\frac{1}{2}\times 2\times 4=4\) Hence Option B is correct. This also equals the area under the line segment from \(x=0\) to \(x=2\).

❓ Why Other Options Are Incorrect

Option A → Uses incorrect height or base.
Option C → Triangle area does not exceed base × height / 2.
Option D → Gives parallelogram area instead of triangle area.

🎯 Strategy

Used

Substitution

Application:

Apply standard triangle area formula using coordinates.

Final Logic:

Half of \(2\times 4\) equals 4.

🧠 Memory Trick

"Triangle area = half base × height"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Area of Triangle

11 Assertion (A): Changing the order of points in a determinant may change the sign, but not the absolute value of the geometric area.
Reason (R): Interchanging any two rows of a determinant changes its sign.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Row interchange changes determinant sign Absolute value removes sign effect Geometric area remains unchanged

✅ Correct Answer Explanation

Changing the order of points interchanges determinant rows, which changes the sign of the determinant. However, geometric area depends on the absolute value of the determinant, so the area magnitude remains unchanged. Thus both Assertion and Reason are true, and the Reason correctly explains the Assertion. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are standard determinant properties and therefore true.
Option B → The Reason is correct because row interchange reverses determinant sign.
Option D → Assertion is true since geometric area uses absolute value.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate determinant sign-change property to area formula involving modulus.

Final Logic:

Sign changes under row interchange, but absolute area stays fixed.

🧠 Memory Trick

"Swap rows → sign flips, area stays"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Properties of Determinants

12 For a triangle with vertices \(\left(k,\ 0\right)\), \(\left(2,\ 0\right)\), \(\left(0,\ 2\right)\), the area is
\(Area=\frac{1}{2}∣2k-4∣\)
Arrange the values of \(k=0,1,2,4\) in increasing order of the area.

A 2, 1, 0, 4

B 1, 2, 0, 4

C 2, 0, 1, 4

D 2, 1, 4, 0

💡 Short Explanation

Compute area for each value of \(k\) Compare absolute values carefully Arrange from smallest to largest

✅ Correct Answer Explanation

Using: \(Area=\frac{1}{2}∣2k-4∣\) For: \(k=0\Rightarrow 2k=1\Rightarrow 1k=2\Rightarrow 0k=4\Rightarrow 2\) Increasing order: \(2, 1, 0, 4\) Hence Option A is correct. The provided answer D is incorrect.

❓ Why Other Options Are Incorrect

Option B → Places area for \(k=1\) before the minimum-area case \(k=2\).
Option C → Incorrectly orders equal-area cases.
Option D → Does not follow increasing numerical area values.

🎯 Strategy

Used

Substitution

Application:

Substitute each value of \(k\) into the area expression and compare results.

Final Logic:

Areas obtained are \(0,1,2,2\), giving order \(2,1,0,4\).

🧠 Memory Trick

"Plug values, then sort areas"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Area of Triangle

📄 Passage

A determinant of order 3 can be expanded along any row or column, and all methods give the same value. For efficiency, expansion should be done along the row or column containing maximum zeros.

If we expand a determinant along the second column instead of the first row, the value of the determinant (and hence area) will:

A Become negative

B Remain the same

C Become zero

D Be halved

💡 Short Explanation

Determinant value is unique Expansion method does not alter value Different rows only change calculation route

✅ Correct Answer Explanation

A determinant has a unique numerical value regardless of the row or column used for expansion. Expanding along the second column or first row only changes the computation method, not the final determinant or area. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Expansion choice alone does not force a sign change.
Option C → Determinant becomes zero only for special matrices, not due to expansion choice.
Option D → Expansion method never halves determinant value.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the passage statement that all expansions yield the same determinant.

Final Logic:

Changing expansion row or column does not alter determinant value.

🧠 Memory Trick

"Different route, same determinant"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Expansion Along Rows and Columns

📄 Passage

A determinant of order 3 can be expanded along any row or column, and all methods give the same value. For efficiency, expansion should be done along the row or column containing maximum zeros.

For easier calculation, expansion should be done along the row or column that:

A Contains maximum zeros

B Contains largest values

C Contains no zeros

D Is always the first row

💡 Short Explanation

Zero terms vanish during expansion Fewer terms simplify calculations Maximum zeros reduce effort

✅ Correct Answer Explanation

During determinant expansion, terms containing zero contribute nothing. Therefore choosing a row or column with maximum zeros reduces the number of calculations. The passage directly states this efficiency rule. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Large numerical values do not simplify determinant expansion.
Option C → Rows without zeros require maximum calculations.
Option D → Any row or column may be chosen; first row is not compulsory.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the exact efficiency rule stated in the passage.

Final Logic:

More zeros imply fewer nonzero expansion terms.

🧠 Memory Trick

"More zeros = easier determinant"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Simplification of Determinants

15 Find the values of \(k\) if the area of triangle with vertices \(\left(k,\ 0\right)\), \(\left(4,\ 0\right)\), \(\left(0,\ 2\right)\)is 4 sq units.

A 2,4

B 0,4

C 0,8

D 2,8

💡 Short Explanation

Use determinant area formula Solve resulting modulus equation Two valid values satisfy area condition

✅ Correct Answer Explanation

Area: \(\frac{1}{2}∣2k-8∣=4∣2k-8∣=82k-8=\pm 8\) Thus: \(k=0 or 8\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Neither pair satisfies the required area equation completely.
Option B → \(k=4\) produces zero area, not 4.
Option D → \(k=2\) gives area 2, not 4.

🎯 Strategy

Used

Substitution

Application:

Apply coordinate-area formula and solve absolute value equation.

Final Logic:

Only \(k=0\) and \(k=8\) satisfy the area condition.

🧠 Memory Trick

"Area equation gives two roots"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Area-Based Variable Finding

16 Find the equation of the line joining \(\left(1,\ 2\right)\)and \(\left(3,\ 6\right)\)using determinants.

A \(y=x+1\)

B \(y=3x\)

C \(y=2x\)

D \(y=2x\)

💡 Short Explanation

Compute slope using two points Slope equals 2 Line passes through origin relation

✅ Correct Answer Explanation

Slope: \(m=\frac{6-2}{3-1}=2\) Using point-slope form: \(y-2=2(x-1)y=2x\) Thus the correct equation is \(y=2x\). Options C and D are identical, but the first correct occurrence is Option C.

❓ Why Other Options Are Incorrect

Option A → Gives slope 1, not 2.
Option B → Gives slope 3, inconsistent with given points.
Option D → Same as Option C; duplicate option formatting issue exists.

🎯 Strategy

Used

Substitution

Application:

Use two-point slope formula and substitute into line equation.

Final Logic:

Points satisfy the equation \(y=2x\).

🧠 Memory Trick

"Rise 4, run 2 → slope 2"

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Equation of Line Through Two Points

17 If
\(∣\begin{pmatrix}1 & 2 & 1\\ 2 & 4 & 1\\ 3 & 6 & 1\end{pmatrix}∣\)
represents three points, what is its value?

A 1

B 2

C 0

D −1

💡 Short Explanation

First two columns are proportional Points are collinear Determinant becomes zero

✅ Correct Answer Explanation

The second column is twice the first: \(\left(2,4,6)=2(1,2,3\right)\) Thus columns are linearly dependent, making the determinant zero. Geometrically, the points are collinear. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Nonzero determinant contradicts column dependence.
Option B → Collinear points cannot produce nonzero area determinant.
Option D → Negative value is impossible because determinant collapses exactly to zero.

🎯 Strategy

Used

Odd One Out

Application:

Identify proportional columns to detect determinant zero instantly.

Final Logic:

Linearly dependent columns always give determinant zero.

🧠 Memory Trick

"Proportional columns → determinant zero"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Properties of Determinants

18 If three points are collinear, then:

A Determinant >0

B Determinant <0

C Determinant =1

D Determinant =0

💡 Short Explanation

Collinear points form zero-area triangle Area determinant becomes zero Standard collinearity condition

✅ Correct Answer Explanation

For three collinear points, the area of the triangle formed is zero. Since triangle area is proportional to the determinant, \(Determinant=0\) Hence Option D is correct. Nonzero determinants indicate non-collinear points.

❓ Why Other Options Are Incorrect

Option A → Positive determinant implies nonzero oriented area.
Option B → Negative determinant also indicates nonzero area with reversed orientation.
Option C → Determinant value 1 does not imply collinearity.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the standard determinant condition for collinear points.

Final Logic:

Zero area directly implies zero determinant.

🧠 Memory Trick

"Collinear ⇒ zero determinant"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Collinearity Condition

19 Find the value of \(x\) if
\(∣\begin{pmatrix}1 & x\\ 2 & 4\end{pmatrix}∣=0.\)

A 2

B 1

C 4

D 0

💡 Short Explanation

Apply \(ad-bc\) formula Solve linear equation Determinant becomes zero at \(x=2\)

✅ Correct Answer Explanation

Evaluate determinant: \(1(4)-2x=04-2x=0x=2\) Thus Option A is correct. Other values fail to satisfy the zero-determinant condition.

❓ Why Other Options Are Incorrect

Option B → Gives determinant \(2\), not zero.
Option C → Gives determinant \(-4\), not zero.
Option D → Gives determinant \(4\), not zero.

🎯 Strategy

Used

Substitution

Application:

Use the determinant formula and solve the resulting equation.

Final Logic:

Only \(x=2\) makes the determinant vanish.

🧠 Memory Trick

"Zero determinant → solve equation"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Determinant Equations

20 A square matrix is said to be singular if:

A It has an inverse

B Its determinant is zero

C It is identity

D It represents a triangle

💡 Short Explanation

Singular matrices are non-invertible Determinant condition defines singularity Zero determinant implies no inverse

✅ Correct Answer Explanation

A square matrix is singular when its determinant equals zero. Such matrices do not possess inverses. Therefore Option B is correct. Identity matrices are non-singular because their determinant equals 1.

❓ Why Other Options Are Incorrect

Option A → Matrices having inverses are non-singular.
Option C → Identity matrices always have determinant 1.
Option D → Singularity is unrelated to geometric triangle representation.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Recall the standard algebraic definition of singular matrices.

Final Logic:

Zero determinant directly defines matrix singularity.

🧠 Memory Trick

"Singular means determinant zero"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Singular and Non-Singular Matrices

CUET UG Mathematics Booster Test 1 - Applications of Determinants in Coordinate Geometry

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