RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
Coordinate geometry turns geometric shapes into algebra you can compute with, and the straight line is the simplest shape to start that translation. This chapter covers slope, the several standard forms of a line's equation, and the formula for distance from a point to a line — tools that let you find angles, check collinearity, and locate a triangle's special points using nothing but arithmetic. Everything here sets up conic sections in the very next chapter.
The slope of the line joining the points (2, -3) and (6, 5) is:
Answer
The slope is 2. Slope m = (y₂-y₁)/(x₂-x₁) = (5-(-3))/(6-2) = 8/4 = 2.
The line 3x + 4y - 12 = 0 has x-intercept and y-intercept respectively:
Answer
The intercepts are 4 and 3. Rewrite as 3x + 4y = 12, then divide by 12: x/4 + y/3 = 1. Comparing with x/a + y/b = 1 gives a = 4 (x-intercept) and b = 3 (y-intercept).
The distance of the point (3, -1) from the line 3x - 4y + 5 = 0 is:
Answer
The distance is 18/5. Using d = |Ax₁+By₁+C|/√(A²+B²) with A=3, B=-4, C=5, x₁=3, y₁=-1: d = |3(3) - 4(-1) + 5| / √(9+16) = |9+4+5|/5 = 18/5.
The equation of the line with slope -3 and y-intercept 7 is:
Answer
3x + y - 7 = 0 is correct. Using slope-intercept form y = mx + c with m = -3, c = 7: y = -3x + 7. Rearranging: 3x + y - 7 = 0.
Assertion (A): The lines 2x + 3y - 6 = 0 and 4x + 6y + 9 = 0 are parallel. Reason (R): Two lines Ax+By+C₁=0 and A'x+B'y+C₂=0 are parallel when A/A' = B/B'.
Answer
Both A and R are true and R is the correct explanation of A. For the two lines, A=2,B=3 and A'=4,B'=6. Check A/A' = 2/4 = 1/2 and B/B' = 3/6 = 1/2 — equal, so the lines are parallel (and since C's ratio 6/-9 differs, they are distinct, not the same line). Assertion A is true. Reason R states exactly this coefficient-ratio test for parallelism, and applying it is how we confirmed A, so R correctly explains A.
Find the equation of the line passing through (1, 2) and (4, 8).
Answer
Slope m = (8-2)/(4-1) = 6/3 = 2. Using point-slope form with (1,2): y - 2 = 2(x - 1). Simplify: y - 2 = 2x - 2, so y = 2x, or 2x - y = 0.
Show that the points A(1, 2), B(3, 6) and C(5, 10) are collinear by finding the equation of line AB and checking C lies on it.
Answer
Slope of AB = (6-2)/(3-1) = 4/2 = 2. Equation of AB using point A(1,2): y - 2 = 2(x-1), giving y = 2x. Check C(5,10): does 10 = 2×5? Yes, 10 = 10. Since C satisfies the equation of line AB, all three points are collinear.
Find the equation of the line passing through (2, 3) and perpendicular to the line 4x - 2y + 5 = 0. Also find the distance of the origin from this new line.
Answer
Step 1 — Find slope of given line: 4x - 2y + 5 = 0 gives y = 2x + 5/2, so slope = 2. Step 2 — The perpendicular line has slope m such that 2×m = -1, so m = -1/2. Step 3 — Equation through (2,3) with slope -1/2: y - 3 = -1/2(x - 2). Multiply by 2: 2y - 6 = -(x-2) = -x + 2. So x + 2y - 8 = 0. Step 4 — Distance of origin (0,0) from x + 2y - 8 = 0: d = |0+0-8|/√(1+4) = 8/√5. So the required line is x + 2y - 8 = 0 and its distance from the origin is 8/√5.
Find the equations of the lines through the point (3, -2) which make an angle of 45° with the line 6x + 5y - 8 = 0.
Answer
Step 1 — Slope of given line: 6x+5y-8=0 gives y = -6x/5 + 8/5, so m₁ = -6/5. Step 2 — Let the required line have slope m. Using tan 45° = |(m-m₁)/(1+mm₁)| = 1: (m + 6/5)/(1 - 6m/5) = ±1. Step 3 — Case +1: m + 6/5 = 1 - 6m/5, so m + 6m/5 = 1 - 6/5, giving 11m/5 = -1/5, so m = -1/11. Case -1: m + 6/5 = -1 + 6m/5, so m - 6m/5 = -1 - 6/5, giving -m/5 = -11/5, so m = 11. Step 4 — Line 1 through (3,-2) with m=-1/11: y+2 = -1/11(x-3) → 11y+22 = -x+3 → x+11y+19=0. Line 2 through (3,-2) with m=11: y+2 = 11(x-3) → y+2 = 11x-33 → 11x - y - 35 = 0. So the two lines are x + 11y + 19 = 0 and 11x - y - 35 = 0.
Read the following and answer the questions that follow: A drone starts at point A(0, 0) and flies in a straight line to point B(8, 6), where distances are measured in metres on a coordinate grid. (a) Find the slope of the drone's flight path AB. (b) Find the equation of the line AB. (c) A charging tower is located at point P(4, 0). Find the perpendicular distance from P to line AB. (d) Find the length of path AB.
Answer
(a) Slope = (6-0)/(8-0) = 6/8 = 3/4. (b) Using y = mx through origin: y = (3/4)x, or 3x - 4y = 0. (c) Distance of P(4,0) from 3x-4y=0: d = |3(4)-4(0)|/√(9+16) = 12/5 = 2.4 m. (d) Length AB = √[(8-0)²+(6-0)²] = √(64+36) = √100 = 10 m.
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