Introduction & Basics — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Introduction & Basics 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 — Introduction & Basics · easy · numerical
The distance between the points $(3, -2)$ and $(-5, 4)$ is:
A. $2\sqrt{7}$
B. $6\sqrt{2}$
C. $10$ ✓ Correct
D. $8$
Solution: $d = \sqrt{(-5-3)^2 + (4+2)^2} = \sqrt{64+36} = 10$.
Q2 — Introduction & Basics · easy · numerical
The distance between the points $(6, 8)$ and $(2, 5)$ is:
A. $7$
B. $5$ ✓ Correct
C. $\sqrt{7}$
D. $25$
Solution: $d = \sqrt{(6-2)^2 + (8-5)^2} = \sqrt{16+9} = 5$.
Q3 — Introduction & Basics · easy · numerical
The midpoint of the segment joining $(-6, 3)$ and $(2, -9)$ is:
A. $(-2, -3)$ ✓ Correct
B. $(-2, 3)$
C. $(-4, -6)$
D. $(2, -3)$
Solution: Midpoint $= \left(\dfrac{-6+2}{2}, \dfrac{3-9}{2}\right) = (-2, -3)$.
Q4 — Introduction & Basics · easy · numerical
The point which divides the segment joining $A(-1, 7)$ and $B(4, -3)$ internally in the ratio $2 : 3$ is:
A. $(1, -3)$
B. $(1, 3)$ ✓ Correct
C. $(2, 1)$
D. $(3, 1)$
Solution: By the section formula, $P = \left(\dfrac{2(4)+3(-1)}{5}, \dfrac{2(-3)+3(7)}{5}\right) = (1, 3)$.
Q5 — Introduction & Basics · easy · numerical
The centroid of the triangle with vertices $(5, 2)$, $(-2, 6)$ and $(6, 7)$ is:
A. $(3, 5)$ ✓ Correct
B. $(3, -5)$
C. $(5, 3)$
D. $(9, 15)$
Solution: Centroid $= \left(\dfrac{5-2+6}{3}, \dfrac{2+6+7}{3}\right) = (3, 5)$.
Q6 — Introduction & Basics · medium · numerical
If the distance between the points $(x, 6)$ and $(1, 2)$ is $5$, the positive value of $x$ is:
A. $4$ ✓ Correct
B. $2$
C. $-2$
D. $6$
Solution: $(x-1)^2 + (6-2)^2 = 25 \Rightarrow (x-1)^2 = 9 \Rightarrow x = 4$ or $x = -2$; the positive value is $4$.
Q7 — Introduction & Basics · medium · numerical
The point on the x-axis which is equidistant from $(7, 6)$ and $(3, 4)$ is:
A. $(5, 0)$
B. $\left(0, \dfrac{15}{2}\right)$
C. $\left(-\dfrac{15}{2}, 0\right)$
D. $\left(\dfrac{15}{2}, 0\right)$ ✓ Correct
Solution: For $(x, 0)$: $(x-7)^2 + 36 = (x-3)^2 + 16 \Rightarrow -14x + 85 = -6x + 25 \Rightarrow x = \dfrac{15}{2}$.
Q8 — Introduction & Basics · medium · numerical
Two vertices of a triangle are $A(3, -5)$ and $B(-7, 4)$, and its centroid is $G(2, -1)$. The third vertex is:
A. $(-2, 10)$
B. $(6, 0)$
C. $(2, -4)$
D. $(10, -2)$ ✓ Correct
Solution: $C = 3G - A - B = (6 - 3 + 7, \ -3 + 5 - 4) = (10, -2)$.
Q9 — Introduction & Basics · medium · numerical
The point which divides the segment joining $(5, -2)$ and $(9, 6)$ externally in the ratio $3 : 1$ is:
A. $(11, 10)$ ✓ Correct
B. $(11, -10)$
C. $(8, 4)$
D. $(10, 11)$
Solution: External division: $\left(\dfrac{3(9) - 1(5)}{3-1}, \dfrac{3(6) - 1(-2)}{3-1}\right) = (11, 10)$.
Q10 — Introduction & Basics · medium · numerical
The ratio in which the point $(-4, 6)$ divides the segment joining $A(-6, 10)$ and $B(3, -8)$ is:
A. $7 : 2$
B. $2 : 7$ ✓ Correct
C. $3 : 7$
D. $2 : 5$
Solution: Let the ratio be $k : 1$; then $\dfrac{3k - 6}{k+1} = -4 \Rightarrow 7k = 2 \Rightarrow k = \dfrac{2}{7}$, i.e. the ratio is $2 : 7$ (the y-coordinate confirms it).
Q11 — Introduction & Basics · medium · numerical
The x-axis divides the segment joining $(2, -3)$ and $(5, 6)$ in the ratio:
A. $1 : 3$
B. $2 : 1$
C. $1 : 2$ ✓ Correct
D. $3 : 2$
Solution: On the x-axis $y = 0$: $\dfrac{6k - 3}{k+1} = 0 \Rightarrow k = \dfrac{1}{2}$, so the ratio is $1 : 2$.
Q12 — Introduction & Basics · medium · numerical
The y-axis divides the segment joining $(-8, 3)$ and $(4, 6)$ in the ratio:
A. $2 : 1$ ✓ Correct
B. $1 : 2$
C. $2 : 3$
D. $3 : 1$
Solution: On the y-axis $x = 0$: $\dfrac{4k - 8}{k+1} = 0 \Rightarrow k = 2$, so the ratio is $2 : 1$.
Q13 — Introduction & Basics · medium · numerical
The area (in square units) of the triangle with vertices $(2, 7)$, $(5, 1)$ and $(-1, 3)$ is:
A. $30$
B. $13$
C. $15$ ✓ Correct
D. $10$
Solution: Area $= \dfrac{1}{2}|2(1-3) + 5(3-7) + (-1)(7-1)| = \dfrac{1}{2}|-4 - 20 - 6| = 15$.
Q14 — Introduction & Basics · medium · numerical
If the points $(k, -2)$, $(3, 0)$ and $(7, 8)$ are collinear, then $k$ equals:
A. $-2$
B. $1$
C. $4$
D. $2$ ✓ Correct
Solution: Collinearity needs area $0$: $k(0-8) + 3(8+2) + 7(-2-0) = -8k + 16 = 0 \Rightarrow k = 2$.
Q15 — Introduction & Basics · hard · numerical
If the area of the triangle with vertices $(-2, k)$, $(3, 1)$ and $(6, 5)$ is $10$ square units, the integer value of $k$ is:
A. $3$
B. $-1$
C. $-3$
D. $1$ ✓ Correct
Solution: $\dfrac{1}{2}|{-2}(1-5) + 3(5-k) + 6(k-1)| = 10 \Rightarrow |17 + 3k| = 20 \Rightarrow k = 1$ (the other root $-\dfrac{37}{3}$ is not an integer).
Q16 — Introduction & Basics · hard · numerical
The midpoints of the sides of a triangle are $(3, 6)$, $(5, 0)$ and $(-2, 2)$. The vertex of the triangle opposite the side whose midpoint is $(3, 6)$ is:
A. $(0, -4)$ ✓ Correct
B. $(10, 4)$
C. $(0, 4)$
D. $(-4, 8)$
Solution: The vertex opposite a midpoint equals the sum of the other two midpoints minus that midpoint: $(5 + (-2) - 3, \ 0 + 2 - 6) = (0, -4)$.
Q17 — Introduction & Basics · hard · numerical
The area (in square units) of the quadrilateral with vertices $(-5, 7)$, $(-4, -5)$, $(-1, -6)$ and $(4, 5)$ taken in order is:
A. $72$ ✓ Correct
B. $36$
C. $144$
D. $65$
Solution: Splitting along the diagonal joining $(-5, 7)$ and $(-1, -6)$ gives triangles of areas $\dfrac{35}{2}$ and $\dfrac{109}{2}$, so the total is $\dfrac{144}{2} = 72$.
Q18 — Introduction & Basics · hard · numerical
A point $P$ on the y-axis is such that the triangle formed by $P$, $A(4, 1)$ and $B(-2, 7)$ has area $12$ square units. The sum of all possible values of the y-coordinate of $P$ is:
A. $9$
B. $1$
C. $10$ ✓ Correct
D. $8$
Solution: For $P(0, p)$: $\dfrac{1}{2}|4(7-p) - 2(p-1)| = \dfrac{1}{2}|30 - 6p| = 12 \Rightarrow p = 1$ or $p = 9$, so the sum is $10$.
Q19 — Introduction & Basics · hard · numerical
The area of the triangle whose vertices are $(a, b+c)$, $(b, c+a)$ and $(c, a+b)$ is:
A. $a+b+c$
B. $\dfrac{1}{2}(a+b+c)$
C. $abc$
D. $0$ ✓ Correct
Solution: Each vertex satisfies $x + y = a + b + c$, so the three points are collinear and the area is $0$.
Q20 — Introduction & Basics · hard · numerical
A triangle has vertices $A(2, -2)$, $B(8, 4)$ and a third vertex $C$ on the x-axis. If the area of the triangle is $15$ square units and $C$ has a positive x-coordinate, then $C$ is:
A. $(-1, 0)$
B. $(0, 9)$
C. $(5, 0)$
D. $(9, 0)$ ✓ Correct
Solution: For $C(c, 0)$: $\dfrac{1}{2}|2(4-0) + 8(0+2) + c(-2-4)| = \dfrac{1}{2}|24 - 6c| = 15 \Rightarrow c = 9$ or $c = -1$; the positive x-coordinate gives $(9, 0)$.