RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Once a point is written as a pair of numbers, geometry becomes arithmetic: you can measure a length, find a midpoint or split a segment in a chosen ratio without touching a ruler. Two formulas carry this whole chapter — the distance formula and the section formula — and everything else, from proving a quadrilateral is a square to locating a point on an axis, is an application of them. Sketch the points roughly before you compute; it catches sign errors instantly.
The distance between the points (2, -3) and (10, 3) is:
Answer
The distance is 10 units. Using the distance formula, d = √[(10 - 2)² + (3 - (-3))²]. d = √[8² + 6²] = √(64 + 36) = √100 = 10. The pair (8, 6) is a scaled 3-4-5 triple, which is why the answer is a whole number.
The point which divides the line segment joining A(1, 2) and B(6, 12) internally in the ratio 2 : 3 is:
Answer
The point is (3, 6). With m : n = 2 : 3, x = (mx₂ + nx₁)/(m + n) = (2 × 6 + 3 × 1)/5 = (12 + 3)/5 = 3. y = (my₂ + ny₁)/(m + n) = (2 × 12 + 3 × 2)/5 = (24 + 6)/5 = 6. So the required point is (3, 6). Note that (3.5, 7) would be the midpoint, i.e. the ratio 1 : 1.
If A(3, -2) and B(-1, 6) are the end points of a diameter of a circle, the centre of the circle is:
Answer
The centre is (1, 2). The centre of a circle is the midpoint of any diameter. Midpoint = ((3 + (-1))/2, (-2 + 6)/2) = (2/2, 4/2) = (1, 2). As a check, the distance from (1, 2) to A(3, -2) is √(4 + 16) = √20, and to B(-1, 6) it is √(4 + 16) = √20 — equal radii.
The point on the x-axis that is equidistant from (2, -5) and (-2, 9) is:
Answer
The point is (-7, 0). Any point on the x-axis is P(x, 0). Equating the squares of the distances: (x - 2)² + (0 + 5)² = (x + 2)² + (0 - 9)² x² - 4x + 4 + 25 = x² + 4x + 4 + 81 -4x + 25 = 4x + 81 -8x = 56, so x = -7. The required point is (-7, 0). Note that (0, -7) lies on the y-axis, not the x-axis.
Assertion (A): The points O(0, 0), P(3, 4) and Q(6, 8) are collinear. Reason (R): Three points are collinear if the sum of the distances between two pairs of them equals the distance between the remaining pair.
Answer
Both A and R are true and R is the correct explanation of A. Reason R is a valid collinearity test. Compute: OP = √(9 + 16) = 5, PQ = √[(6 - 3)² + (8 - 4)²] = √(9 + 16) = 5, and OQ = √(36 + 64) = √100 = 10. Since OP + PQ = 5 + 5 = 10 = OQ, the point P lies on the segment OQ, so the three points are collinear. Assertion A is therefore true and follows directly from the test in R.
Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer
By the distance formula, PQ² = (10 - 2)² + (y + 3)² = 100. 64 + (y + 3)² = 100 (y + 3)² = 36, so y + 3 = ±6. Taking y + 3 = 6 gives y = 3; taking y + 3 = -6 gives y = -9. Hence y = 3 or y = -9.
Find the ratio in which the point (-4, 6) divides the line segment joining A(-6, 10) and B(3, -8).
Answer
Let the required ratio be k : 1. Using the section formula for the x-coordinate: -4 = (k × 3 + 1 × (-6))/(k + 1) -4(k + 1) = 3k - 6 -4k - 4 = 3k - 6 2 = 7k, so k = 2/7. The ratio is 2/7 : 1, that is 2 : 7. Verification with the y-coordinate: (2 × (-8) + 7 × 10)/(2 + 7) = (-16 + 70)/9 = 54/9 = 6 ✔
Find the coordinates of the point which divides the line segment joining (-2, 3) and (6, -5) internally in the ratio 3 : 1.
Answer
Using the section formula with m = 3 and n = 1: x = (3 × 6 + 1 × (-2))/(3 + 1) = (18 - 2)/4 = 16/4 = 4. y = (3 × (-5) + 1 × 3)/(3 + 1) = (-15 + 3)/4 = -12/4 = -3. The required point is (4, -3).
Show that the points A(1, 2), B(5, 4), C(3, 8) and D(-1, 6), taken in order, are the vertices of a square.
Answer
Step 1 — Find the four side lengths. AB = √[(5 - 1)² + (4 - 2)²] = √(16 + 4) = √20 BC = √[(3 - 5)² + (8 - 4)²] = √(4 + 16) = √20 CD = √[(-1 - 3)² + (6 - 8)²] = √(16 + 4) = √20 DA = √[(1 - (-1))² + (2 - 6)²] = √(4 + 16) = √20 All four sides are equal, so ABCD is a rhombus. Step 2 — Find the two diagonals. AC = √[(3 - 1)² + (8 - 2)²] = √(4 + 36) = √40 BD = √[(-1 - 5)² + (6 - 4)²] = √(36 + 4) = √40 The diagonals are equal. Step 3 — A quadrilateral with all four sides equal and both diagonals equal is a square. Hence ABCD is a square. Step 4 — Consistency check: for a square of side √20 the diagonal should be √20 × √2 = √40, which matches exactly.
(a) Find the point on the y-axis which is equidistant from the points A(6, 5) and B(-4, 3). (b) Find the coordinates of the points of trisection of the line segment joining P(2, -2) and Q(-7, 4).
Answer
(a) Any point on the y-axis has the form R(0, y). RA² = RB² (0 - 6)² + (y - 5)² = (0 + 4)² + (y - 3)² 36 + y² - 10y + 25 = 16 + y² - 6y + 9 61 - 10y = 25 - 6y 36 = 4y, so y = 9. The required point is (0, 9). Check: RA = √(36 + 16) = √52 and RB = √(16 + 36) = √52 ✔ (b) The points of trisection divide PQ in the ratios 1 : 2 and 2 : 1. For 1 : 2: x = (1 × (-7) + 2 × 2)/3 = (-7 + 4)/3 = -1, y = (1 × 4 + 2 × (-2))/3 = (4 - 4)/3 = 0. First point: (-1, 0). For 2 : 1: x = (2 × (-7) + 1 × 2)/3 = (-14 + 2)/3 = -4, y = (2 × 4 + 1 × (-2))/3 = (8 - 2)/3 = 2. Second point: (-4, 2). So the points of trisection are (-1, 0) and (-4, 2).
Read the following and answer the questions that follow: A town planner draws a map on graph paper, taking 1 unit as 100 metres. On this map the school is at A(2, 3), the public library is at B(10, 9) and a small park is at C(6, 6). (a) Find the distance on the map between the school and the library. (b) Show that the park lies exactly at the midpoint of the straight road joining the school and the library. (c) A canteen is to be built on the same road so that it divides the journey from the school to the library in the ratio 3 : 1. Find its coordinates. (d) Find the map distance from the park to the library, and state the actual distance in metres.
Answer
(a) AB = √[(10 - 2)² + (9 - 3)²] = √(64 + 36) = √100 = 10 units. (b) Midpoint of AB = ((2 + 10)/2, (3 + 9)/2) = (6, 6), which is exactly the point C. So the park is the midpoint of the road. (c) Using the section formula with m : n = 3 : 1 on A(2, 3) and B(10, 9): x = (3 × 10 + 1 × 2)/4 = 32/4 = 8 y = (3 × 9 + 1 × 3)/4 = 30/4 = 7.5 The canteen is at (8, 7.5). (d) CB = √[(10 - 6)² + (9 - 6)²] = √(16 + 9) = √25 = 5 units, which is half of AB as expected for a midpoint. Since 1 unit represents 100 metres, the actual distance is 5 × 100 = 500 metres.
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