SSC CGL (Prelims) General: Coordinate Geometry questions with solutions

Q1. What is the equation of a line passing through \((0,4)\) with slope \(-2\)?

1. y = -2x + 4
2. y = 2x - 4
3. y = -2x - 4
4. y = 2x + 4

Answer: y = -2x + 4

A line with slope \(m\) and y-intercept \(c\) has equation \(y = mx + c\). Here, slope = -2 and the line passes through \((0,4)\), so \(c = 4\).

Q2. What is the slope of the line perpendicular to $y = 5x - 8$?

1. 5
2. -5
3. 1/5
4. -1/5

Answer: -1/5

The line $y=5x-8$ has slope $5$. A line perpendicular to it has slope equal to the negative reciprocal of 5, which is $-\frac{1}{5}$.

Q3. Which line has slope \(-2\) and passes through \((3,5)\)?

1. y = -2x + 11
2. y = 2x + 11
3. y = -2x - 11
4. y = 2x - 11

Answer: y = -2x + 11

For slope \(m=-2\), the line is \(y=-2x+c\). Substituting point \((3,5)\) gives \(5=-6+c\), so \(c=11\). Therefore the equation is \(y=-2x+11\).

Q4. A line \(L\) is the perpendicular bisector of the line segment joining points \(P(4,6)\) and \(Q(10,2)\). What is the y-intercept of line \(L\)?

1. 4
2. - 6.5
3. 6
4. 7

Answer: - 6.5

The midpoint of \(P(4,6)\) and \(Q(10,2)\) is \((7,4)\). The slope of \(PQ\) is \((2-6)/(10-4)=-4/6=-2/3\), so the perpendicular slope is \(3/2\). Using point-slope form through \((7,4)\), the line is \(y-4=\frac{3}{2}(x-7)\), giving \(y=-\frac{13}{2}=-6.5\) when \(x=0\).

Q5. Find the slope of the line $4x - 5y = 10$.

1. 4/5
2. − 5/4
3. 5/4
4. − 4/5

Answer: 4/5

For a line in the form $Ax + By + C = 0$, the slope is $-A/B$. Here, $4x - 5y = 10$ gives $y = \frac{4}{5}x - 2$. So the slope is $\frac{4}{5}$.

Q6. Find the equation of the perpendicular bisector of the line segment joining the points $(1,5)$ and $(5,9)$.

1. $y=-x+10$
2. $y=x+4$
3. $y=-x+14$
4. $y=x-4$

Answer: $y=-x+10$

The midpoint of $(1,5)$ and $(5,9)$ is $(3,7)$. The slope of the segment is $(9-5)/(5-1)=1$, so the perpendicular slope is $-1$. Using point-slope form through $(3,7)$ gives $y-7=-1(x-3)$, i.e. $y=-x+10$.

Q7. Determine the slope of a straight line that passes through the coordinates \((5,-2)\) and \((8,7)\).

1. 3
2. 1/3
3. -3
4. 2

Answer: 3

The slope of a line through \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{y_2-y_1}{x_2-x_1}\). Here, \(m=\frac{7-(-2)}{8-5}=\frac{9}{3}=3\).

Q8. Find the midpoint of the segment joining \((4,-2)\) and \((10,6)\).

1. (7, 4)
2. (6, 2)
3. (7, 2)
4. (5, 3)

Answer: (7, 2)

The midpoint of two points is found by averaging their coordinates. Here, \(x=\frac{4+10}{2}=7\) and \(y=\frac{-2+6}{2}=2\). So the midpoint is \((7,2)\).

Q9. Triangle PQR has its centroid at \(G(2,-1)\), and vertex \(P\) is located at \((-4,5)\). If point \(M\) is the midpoint of side \(QR\), what are the coordinates of \(M\)?

1. (4, -3)
2. (5, -4)
3. (3, -2)
4. (6, -5)

Answer: (5, -4)

In a triangle, the centroid divides each median in the ratio \(2:1\) from the vertex. So for median \(PM\), \(G\) divides it such that \(G = \frac{2M+P}{3}\). Solving gives \(M=(5,-4)\).

Q10. A line \(y = px - 4\) passes through \((2,6)\). Find \(p\).

1. 3
2. 4
3. 5
4. 6

Answer: 5

Since the point \((2,6)\) lies on the line, it must satisfy the equation. Substituting gives \(6 = 2p - 4\), so \(2p = 10\) and \(p = 5\).

Q11. Find the slope of the line \(3x-5y=10\).

1. 3/5
2. −3/5
3. 5/3
4. −5/3

Answer: 3/5

Rearranging \(3x-5y=10\) gives \(-5y=-3x+10\), so \(y=\frac{3}{5}x-2\). The slope is the coefficient of \(x\), which is \(\frac{3}{5}\).

Q12. The graph of the equation \(2x + 5y = 10\) passes through which point?

1. (5, 0)
2. (0, 2)
3. (-5, 4)
4. All of the above

Answer: All of the above

Substituting \((5,0)\) gives \(2(5)+5(0)=10\), and substituting \((0,2)\) gives \(2(0)+5(2)=10\). But \((-5,4)\) gives \(-10+20=10\), so all listed points satisfy the equation.

Q13. Find the equation of the line passing through the point \((2, 3)\) with a slope of 4.

1. y = 4x - 5
2. y = 4x + 3
3. y = 4x - 11
4. y = 4x + 1

Answer: y = 4x - 5

Using point-slope form, \(y-3=4(x-2)\). Simplifying gives \(y=4x-5\).

Q14. The centroid of triangle ABC is G(2, 4). If A is (5, 2) and B is (-1, 6), find the coordinates of C.

1. (2, 4)
2. (3, 5)
3. (2, 3)
4. (4, 4)

Answer: (2, 4)

For a triangle with vertices A(x1,y1), B(x2,y2), C(x3,y3), the centroid is \(\left(\frac{x1+x2+x3}{3},\frac{y1+y2+y3}{3}\right)\). Using G(2,4), A(5,2), and B(-1,6), solve for C. The coordinates come out to (2,4).

Q15. Which point lies on the line $y = 2x - 5$?

1. (3, 1)
2. (4, 3)
3. (5, 5)
4. (3, 1), (4, 3) and (5, 5) all

Answer: (3, 1), (4, 3) and (5, 5) all

Check each option in the equation $y = 2x - 5$. For $(3,1)$, $2(3)-5=1$; for $(4,3)$, $2(4)-5=3$; and for $(5,5)$, $2(5)-5=5$. So all three points satisfy the equation.

Q16. A right triangle has vertices at \(P(0,0)\), \(Q(10,0)\), and \(R(0,24)\). What are the coordinates of its circumcenter?

1. (5, 12)
2. (12, 5)
3. (10, 24)
4. (0, 0)

Answer: (5, 12)

In a right triangle, the circumcenter lies at the midpoint of the hypotenuse. Here the hypotenuse is between \(Q(10,0)\) and \(R(0,24)\), whose midpoint is \((5,12)\).

Q17. What is the equation of the line passing through \((2,3)\) and \((4,11)\)?

1. y = 4x – 5
2. y = 3x – 2
3. y = 4x + 5
4. y = 2x + 1

Answer: y = 4x – 5

The slope of the line is \((11-3)/(4-2)=8/2=4\). Using point \((2,3)\), the line is \(y-3=4(x-2)\), which simplifies to \(y=4x-5\).

Q18. If a line passing through \((2,-3)\) has a slope of 5, find its y-intercept.

1. – 13
2. 13
3. – 10
4. 10

Answer: – 13

For a line with slope 5, the equation is \(y = 5x + c\). Since it passes through \((2,-3)\), substitute to get \(-3 = 5(2) + c\), so \(c = -13\). Thus, the y-intercept is -13.

Q19. In a triangle with vertices \(A(x,y)\), \(B(6,-2)\), and \(C(3,8)\), the centroid is at \(G(5,3)\). What are the coordinates of vertex \(A\)?

1. (5, 2)
2. (6, 3)
3. (7, 4)
4. (8, 5)

Answer: (6, 3)

The centroid of a triangle is \(\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\). Using \(G(5,3)\) with \(B(6,-2)\) and \(C(3,8)\), we get \(x+6+3=15\Rightarrow x=6\) and \(y-2+8=9\Rightarrow y=3\). So \(A=(6,3)\).

Q20. What is the minimum distance from the origin $(0,0)$ to a point on the line segment connecting the points $A(6,0)$ and $B(0,8)$?

1. 3.6
2. 4.2
3. 4.8
4. 5.4

Answer: 4.8

The line through $(6,0)$ and $(0,8)$ is $\frac{x}{6}+\frac{y}{8}=1$, or $4x+3y=24$. The distance from the origin to this line is $\frac{|24|}{\sqrt{4^2+3^2}}=\frac{24}{5}=4.8$. Since the perpendicular foot lies on the segment, this is the minimum distance.

Q21. Find the slope of the line passing through the points $(-3, 4)$ and $(5, -2)$.

1. 3/4
2. – 3/4
3. 4/3
4. – 4/3

Answer: – 3/4

The slope of a line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$. Here, $m=\frac{-2-4}{5-(-3)}=\frac{-6}{8}=-\frac{3}{4}$.

Q22. Find the y-intercept of the line $3x + 4y = 12$.

1. 12
2. 4
3. 3
4. 7

Answer: 3

The y-intercept is found by setting $x=0$. Then $4y=12$, so $y=3$. Hence, the y-intercept is 3.

Q23. What is the length of the segment joining \((2,3)\) and \((8,11)\)?

1. 8
2. 9
3. 12
4. 10

Answer: 10

The distance between \((2,3)\) and \((8,11)\) is \(\sqrt{(8-2)^2+(11-3)^2}=\sqrt{6^2+8^2}=\sqrt{100}=10\).

Q24. Are the lines \(3x-4y=12\) and \(4x+3y=7\) perpendicular?

1. Yes
2. No
3. Only if they intersect at the y-axis
4. Cannot be determined

Answer: Yes

For \(3x-4y=12\), slope \(m_1=\frac{3}{4}\). For \(4x+3y=7\), slope \(m_2=-\frac{4}{3}\). Since \(m_1m_2=\frac{3}{4}\cdot\left(-\frac{4}{3}\right)=-1\), the lines are perpendicular.

Q25. Find the point where the line y = 3x - 12 crosses the x-axis.

1. (0, -12)
2. (-4, 0)
3. (0, 4)
4. (4, 0)

Answer: (4, 0)

A line crosses the x-axis where y = 0. Putting y = 0 in y = 3x - 12 gives 0 = 3x - 12, so x = 4. Hence the point is (4, 0).

Q26. A triangle has vertices at X(1,2), Y(4,6), and Z(-3,5). What is its area?

1. 10.5 square units
2. 12.5 square units
3. 11.5 square units
4. 14.5 square units

Answer: 12.5 square units

Using the area formula for coordinates, \(\frac12 |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\), we get \(\frac12 |1(6-5)+4(5-2)+(-3)(2-6)|\). This becomes \(\frac12 |1+12+12| = \frac{25}{2} = 12.5\).

Q27. In triangle ABC, an altitude AD is drawn from point A to side BC. Given the coordinates A(1, 5), B(2, 1), and C(6, 3), what is the slope of the line representing altitude AD?

1. 2
2. -2
3. -1/2
4. 1/2

Answer: -2

The slope of BC is \((3-1)/(6-2)=2/4=1/2\). Since AD is an altitude, it is perpendicular to BC, so its slope is the negative reciprocal of \(1/2\), which is \(-2\).

Q28. Find the x-intercept of \(3x - 6y = 12\).

1. 3
2. 4
3. 5
4. -2

Answer: 4

The x-intercept is found by setting \(y=0\). Then \(3x - 6(0) = 12\), so \(3x = 12\) and \(x = 4\).

Q29. The line \(3x+2y=12\) passes through which of the following points?

1. (2, 3)
2. (4, 1)
3. (1, 6)
4. (1, 4)

Answer: (2, 3)

A point lies on the line if its coordinates satisfy the equation. Substituting \((2,3)\) gives \(3(2)+2(3)=6+6=12\), so it lies on the line.

Q30. Which of the following lines is parallel to the line $4x - 3y = 7$?

1. 4x - 3y = 5
2. 3x - 4y = 7
3. 4x + 3y = 7
4. 2x - 3y = 5

Answer: 4x - 3y = 5

For a line $ax + by = c$, the slope is $-a/b$. Parallel lines must have the same slope, so they must have the same coefficients of x and y. Among the options, only $4x - 3y = 5$ matches the given line's coefficients.