High-Yield JEE Extensions — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics High-Yield JEE Extensions MCQs with step-by-step solutions (20 questions). Part of Straight Lines. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — High-Yield JEE Extensions · easy · numerical
If the pair of lines $kx^2 + 7xy - 4y^2 = 0$ are perpendicular to each other, then $k$ equals:
A. $0$
B. $4$ ✓ Correct
C. $7$
D. $-4$
Solution: A homogeneous pair $ax^2 + 2hxy + by^2 = 0$ is perpendicular when $a + b = 0$: here $k - 4 = 0$, so $k = 4$.
Q2 — High-Yield JEE Extensions · easy · numerical
The angle between the pair of lines $x^2 + 4xy + y^2 = 0$ is:
A. $60^\circ$ ✓ Correct
B. $90^\circ$
C. $45^\circ$
D. $30^\circ$
Solution: $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{|a + b|} = \dfrac{2\sqrt{4 - 1}}{2} = \sqrt{3}$, so $\theta = 60^\circ$.
Q3 — High-Yield JEE Extensions · easy · numerical
The separate equations of the lines given by $2x^2 - 5xy + 2y^2 = 0$ are:
A. $x - y = 0$ and $2x - y = 0$
B. $2x + y = 0$ and $x + 2y = 0$
C. $2x - y = 0$ and $x - 2y = 0$ ✓ Correct
D. $2x - y = 0$ and $x + 2y = 0$
Solution: $2x^2 - 5xy + 2y^2 = (2x - y)(x - 2y)$, so the lines are $2x - y = 0$ and $x - 2y = 0$.
Q4 — High-Yield JEE Extensions · easy · numerical
If the line $(3x + y - 5) + \lambda(x - 3y + 5) = 0$ passes through the point $(0, 2)$, then $\lambda$ equals:
A. $-3$ ✓ Correct
B. $3$
C. $-\dfrac{1}{3}$
D. $\dfrac{1}{3}$
Solution: Substitute $(0, 2)$: $(0 + 2 - 5) + \lambda(0 - 6 + 5) = -3 - \lambda = 0$, so $\lambda = -3$.
Q5 — High-Yield JEE Extensions · easy · numerical
For every value of $\lambda$, the line $(x + y - 2) + \lambda(x - y) = 0$ passes through the fixed point:
A. $(-1, 1)$
B. $(1, 1)$ ✓ Correct
C. $(2, 0)$
D. $(1, -1)$
Solution: The fixed point satisfies both $x + y - 2 = 0$ and $x - y = 0$, giving $(1, 1)$.
Q6 — High-Yield JEE Extensions · medium · numerical
For all values of $k$, the line $(1 + k)x + (k - 4)y - 5k = 0$ passes through the fixed point:
A. $(5, 0)$
B. $(-4, 1)$
C. $(4, 1)$ ✓ Correct
D. $(1, 4)$
Solution: Rewrite as $(x - 4y) + k(x + y - 5) = 0$; the fixed point solves $x = 4y$ and $x + y = 5$, giving $(4, 1)$.
Q7 — High-Yield JEE Extensions · medium · numerical
If the lines $2x + y - 5 = 0$, $x - 3y + 8 = 0$ and $kx + 2y - 9 = 0$ are concurrent, then $k$ equals:
A. $1$
B. $-3$
C. $9$
D. $3$ ✓ Correct
Solution: The first two lines meet at $(1, 3)$; substituting in the third: $k + 6 - 9 = 0$, so $k = 3$.
Q8 — High-Yield JEE Extensions · medium · numerical
If the lines $x + 4y - 2 = 0$, $3x + 2y + 4 = 0$ and $kx - 3y + 7 = 0$ are concurrent, then $k$ equals:
A. $-2$
B. $\dfrac{7}{2}$
C. $2$ ✓ Correct
D. $4$
Solution: The first two lines meet at $(-2, 1)$; substituting in the third: $-2k - 3 + 7 = 0$, so $k = 2$.
Q9 — High-Yield JEE Extensions · medium · numerical
The foot of the perpendicular from the point $(4, 9)$ to the line $3x + 4y - 23 = 0$ is:
A. $(1, -5)$
B. $(1, 5)$ ✓ Correct
C. $(-1, 5)$
D. $(5, 2)$
Solution: $\dfrac{x - 4}{3} = \dfrac{y - 9}{4} = -\dfrac{3(4) + 4(9) - 23}{3^2 + 4^2} = -1$, giving the foot $(1, 5)$, which indeed satisfies $3(1) + 4(5) - 23 = 0$.
Q10 — High-Yield JEE Extensions · medium · numerical
The image of the point $(2, 7)$ in the line $y = x$ is:
A. $(7, 2)$ ✓ Correct
B. $(2, -7)$
C. $(-7, -2)$
D. $(-2, -7)$
Solution: Reflection in the mirror line $y = x$ interchanges the coordinates, so the image is $(7, 2)$.
Q11 — High-Yield JEE Extensions · medium · numerical
The image of the point $(6, 2)$ in the line $3x + 4y - 1 = 0$ is:
A. $(0, -6)$ ✓ Correct
B. $(12, 10)$
C. $(-6, 0)$
D. $(0, 6)$
Solution: $\dfrac{x - 6}{3} = \dfrac{y - 2}{4} = -\dfrac{2(3(6) + 4(2) - 1)}{25} = -2$, giving the image $(0, -6)$; the midpoint $(3, -2)$ lies on the line.
Q12 — High-Yield JEE Extensions · medium · numerical
The angle between the pair of straight lines $2x^2 + 7xy + 3y^2 = 0$ is:
A. $60^\circ$
B. $45^\circ$ ✓ Correct
C. $90^\circ$
D. $30^\circ$
Solution: $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{|a + b|} = \dfrac{2\sqrt{\frac{49}{4} - 6}}{5} = \dfrac{2\cdot\frac{5}{2}}{5} = 1$, so $\theta = 45^\circ$.
Q13 — High-Yield JEE Extensions · medium · numerical
The equation $2x^2 + xy - y^2 + 5x - y + \lambda = 0$ represents a pair of straight lines when $\lambda$ equals:
A. $\dfrac{9}{2}$
B. $-2$
C. $4$
D. $2$ ✓ Correct
Solution: The condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ with $a = 2$, $b = -1$, $h = \tfrac{1}{2}$, $g = \tfrac{5}{2}$, $f = -\tfrac{1}{2}$, $c = \lambda$ gives $-\dfrac{9\lambda}{4} + \dfrac{18}{4} = 0$, so $\lambda = 2$.
Q14 — High-Yield JEE Extensions · medium · numerical
If the sum of the slopes of the lines given by $2x^2 + kxy - 3y^2 = 0$ is $5$, then $k$ equals:
A. $\dfrac{15}{2}$
B. $-15$
C. $15$ ✓ Correct
D. $5$
Solution: For $ax^2 + 2hxy + by^2 = 0$, $m_1 + m_2 = -\dfrac{2h}{b} = -\dfrac{k}{-3} = \dfrac{k}{3}$; setting $\dfrac{k}{3} = 5$ gives $k = 15$.
Q15 — High-Yield JEE Extensions · hard · numerical
A ray of light travelling along the line $x - 2y + 3 = 0$ strikes the $x$-axis and is reflected. The equation of the reflected ray is:
A. $2x - y + 6 = 0$
B. $x + 2y + 3 = 0$ ✓ Correct
C. $x + 2y - 3 = 0$
D. $x - 2y - 3 = 0$
Solution: The ray meets the $x$-axis at $(-3, 0)$; the image of the point $(5, 4)$ of the incident ray in the $x$-axis is $(5, -4)$, and the line through $(-3, 0)$ and $(5, -4)$ has slope $-\dfrac{1}{2}$, giving $x + 2y + 3 = 0$.
Q16 — High-Yield JEE Extensions · hard · numerical
The equation of the bisector of the angle containing the origin, for the lines $4x + 3y - 6 = 0$ and $5x + 12y + 9 = 0$, is:
A. $9x - 7y - 41 = 0$
B. $9x + 7y - 41 = 0$
C. $7x + 9y + 3 = 0$
D. $7x + 9y - 3 = 0$ ✓ Correct
Solution: At the origin the two expressions have opposite signs, so take $\dfrac{4x + 3y - 6}{5} = -\dfrac{5x + 12y + 9}{13}$, which simplifies to $77x + 99y - 33 = 0$, i.e. $7x + 9y - 3 = 0$.
Q17 — High-Yield JEE Extensions · hard · numerical
The equation $9x^2 - 24xy + 16y^2 - 12x + 16y - 12 = 0$ represents a pair of parallel straight lines. The distance between them is:
A. $\dfrac{4}{5}$
B. $8$
C. $\dfrac{8}{5}$ ✓ Correct
D. $2$
Solution: With $u = 3x - 4y$ the equation becomes $u^2 - 4u - 12 = (u - 6)(u + 2) = 0$, giving $3x - 4y - 6 = 0$ and $3x - 4y + 2 = 0$; distance $= \dfrac{|-6 - 2|}{5} = \dfrac{8}{5}$.
Q18 — High-Yield JEE Extensions · hard · numerical
The member of the family of lines $(2x - y - 2) + \lambda(x + 3y - 8) = 0$ that is farthest from the point $(5, 6)$ is:
A. $3x + 4y - 14 = 0$ ✓ Correct
B. $3x + 4y + 14 = 0$
C. $x + 3y - 8 = 0$
D. $4x - 3y - 2 = 0$
Solution: Every member passes through the fixed point $P(2, 2)$; the farthest member is perpendicular to $PQ$ with $Q(5, 6)$, i.e. the line through $P$ with normal $(3, 4)$: $3x + 4y - 14 = 0$.
Q19 — High-Yield JEE Extensions · hard · numerical
Every line $(2x + y - 8) + \lambda(x - 2y + 11) = 0$ passes through a fixed point $P$. The distance of $P$ from the line $4x + 3y + 5 = 0$ is:
A. $\dfrac{17}{5}$
B. $\dfrac{27}{5}$ ✓ Correct
C. $27$
D. $\dfrac{27}{7}$
Solution: $P$ solves $2x + y = 8$ and $x - 2y = -11$, giving $P(1, 6)$; then $d = \dfrac{|4(1) + 3(6) + 5|}{5} = \dfrac{27}{5}$.
Q20 — High-Yield JEE Extensions · hard · numerical
If $y = 3x$ is one of the lines given by $kx^2 + 5xy - 3y^2 = 0$, the other line is:
A. $4x + 3y = 0$ ✓ Correct
B. $3x - 4y = 0$
C. $3x + 4y = 0$
D. $4x - 3y = 0$
Solution: Substituting $y = 3x$: $k + 15 - 27 = 0$, so $k = 12$; then $12x^2 + 5xy - 3y^2 = (3x - y)(4x + 3y)$, so the other line is $4x + 3y = 0$.