IPMPrepFor what value of k are the points (2, −5), (6, −4) and (18, k) collinear?
Solution
Slope through the first two points = 14.
The third point must have the same slope from (2, −5): (k − −5)/(18 − 2) = 14.
k = −1.
Answer: −1
Quantitative Ability · Geometry
Coordinate Geometry: IPMAT practice questions
24 questions, 16 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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Find the perpendicular distance of the point (0, 0) from the line 6x + 8y − 40 = 0.
Solution
Distance = |a x0 + b y0 + c|/√(a2 + b2) = |6(0) + 8(0) − 40|/√(36 + 64).
= |−40|/10 = 4.
Answer: 4
Find the distance between the points (2, 0) and (6, 3).
Solution
√((6 − 2)2 + (3 − 0)2) = √(16 + 9) = 5.
Answer: (A) 5
Find the radius of the circle x2 + y2 − 8x + 12y + 16 = 0.
Solution
Complete the squares: (x − 4)2 + (y + 6)2 = 16 + 36 − 16 = 36.
Radius = √36 = 6.
Answer: 6
The lines 3x + y = 18 and x − y = −2 meet at (h, k). Find h + k.
Solution
Solving the two equations gives x = 4 and y = 6.
h + k = 10.
Answer: (C) 10
Find the radius of the circle x2 + y2 − 2x + 4y + 1 = 0.
Solution
Complete the squares: (x − 1)2 + (y + 2)2 = 1 + 4 − 1 = 4.
Radius = √4 = 2.
Answer: 2
Find the perpendicular distance of the point (1, 2) from the line 3x + 4y + 1 = 0.
Solution
Distance = |a x0 + b y0 + c|/√(a2 + b2) = |3(1) + 4(2) + 1|/√(9 + 16).
= |12|/5 = 2.
Answer: (D) 2
Find the perpendicular distance of the point (−1, −4) from the line 4x + 3y − 21 = 0.
Solution
Distance = |a x0 + b y0 + c|/√(a2 + b2) = |4(−1) + 3(−4) − 21|/√(16 + 9).
= |−37|/5 = 1.
Answer: (A) 1
Find the distance between the points (−1, −3) and (11, −12).
Solution
√((11 − −1)2 + (−12 − −3)2) = √(144 + 81) = 15.
Answer: (B) 15
Find the area of the triangle with vertices (6, −6), (−5, 4) and (−1, 7).
Solution
Area = (12)|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|.
= (12)|6(−3) + −5(13) + −1(−10)| = 36.5.
Answer: 36.5
Find the radius of the circle x2 + y2 + 4x + 6y − 3 = 0.
Solution
Complete the squares: (x + 2)2 + (y + 3)2 = 4 + 9 + 3 = 16.
Radius = √16 = 4.
Answer: 4
Find the area of the triangle with vertices (−2, 3), (−6, 1) and (−4, 7).
Solution
Area = (12)|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|.
= (12)|−2(−6) + −6(4) + −4(2)| = 10.
Answer: (D) 10
Find the distance between the points (−4, 4) and (−11, −20).
Solution
√((−11 − −4)2 + (−20 − 4)2) = √(49 + 576) = 25.
Answer: 25
For what value of k are the points (0, 3), (4, 7) and (−8, k) collinear?
Solution
Slope through the first two points = 1.
The third point must have the same slope from (0, 3): (k − 3)/(−8 − 0) = 1.
k = −5.
Answer: (A) −5
Find the area of the region enclosed by |x| + |y| = 6.
Solution
The curve is a square with vertices (±6, 0) and (0, ±6); its diagonals are 12 each.
Area = (12) × 12 × 12 = 72.
Answer: 72
The lines 2x + 4y = 30 and 2x − 4y = −2 meet at (h, k). Find h + k.
Solution
Solving the two equations gives x = 7 and y = 4.
h + k = 11.
Answer: 11
Find the area of the triangle formed by the line x + 4y = 8 and the coordinate axes.
Solution
x-intercept = 81 = 8; y-intercept = 84 = 2.
Area = (12) × 8 × 2 = 8.
Answer: 8
Find the area of the triangle formed by the line 2x + 5y = 10 and the coordinate axes.
Solution
x-intercept = 102 = 5; y-intercept = 105 = 2.
Area = (12) × 5 × 2 = 5.
Answer: (C) 5
Find the area of the triangle formed by the line x + 2y = 10 and the coordinate axes.
Solution
x-intercept = 101 = 10; y-intercept = 102 = 5.
Area = (12) × 10 × 5 = 25.
Answer: 25
For what value of k are the points (2, 3), (4, 5) and (0, k) collinear?
Solution
Slope through the first two points = 1.
The third point must have the same slope from (2, 3): (k − 3)/(0 − 2) = 1.
k = 1.
Answer: 1
The lines x − y = 6 and 5x + 2y = 23 meet at (h, k). Find h + k.
Solution
Solving the two equations gives x = 5 and y = −1.
h + k = 4.
Answer: 4
Find the area of the region enclosed by 6|x| + |y| = 6.
Solution
The vertices are (±1, 0) and (0, ±6): a rhombus.
Area = (12) × 2 × 12 = 12.
Answer: 12
Find the area of the triangle with vertices (7, −4), (−4, 6) and (7, −3).
Solution
Area = (12)|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|.
= (12)|7(9) + −4(1) + 7(−10)| = 5.5.
Answer: 5.5
Find the area of the region enclosed by |x| + |y| = 4.
Solution
The curve is a square with vertices (±4, 0) and (0, ±4); its diagonals are 8 each.
Area = (12) × 8 × 8 = 32.
Answer: 32
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.