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Exercise 12.1 · Q2

Q.If A(1,4)(1, 4), B(2,−3)(2, -3) and C(−1,−2)(-1, -2) are the vertices of a △ABC\triangle ABC, then find the equation of

(i) the median through A
(ii) the altitude through A
(iii) the perpendicular bisector of BC
Arunachal CbseNCERTSubjective· 3mImportance★★★★★est
5% · 2/40 Questions
✓ Free question

Find the midpoint and slope of BC, then build each requested line using the point-slope form.

Midpoint: (x1+x22,y1+y22)\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right). Slope: m=y2−y1x2−x1m=\dfrac{y_2-y_1}{x_2-x_1}. Perpendicular slope: −1m-\dfrac1m. Point-slope form: y−y0=m(x−x0)y-y_0=m(x-x_0).

  1. Given vertices. A(1,4)A(1,4), B(2,−3)B(2,-3), C(−1,−2)C(-1,-2).

  2. Preliminary: midpoint DD of BCBC (needed for both the median and the perpendicular bisector):

D=(2+(−1)2,−3+(−2)2)=(12,−52)D = \left(\frac{2+(-1)}{2}, \frac{-3+(-2)}{2}\right) = \left(\frac12, -\frac52\right)

  1. Preliminary: slope of BCBC (needed for both the altitude and the perpendicular bisector):

mBC=−2−(−3)−1−2=1−3=−13m_{BC} = \frac{-2-(-3)}{-1-2} = \frac{1}{-3} = -\frac13

  1. (i) Median through A — the line joining A(1,4)A(1,4) to D(12,−52)D\left(\tfrac12,-\tfrac52\right).

mAD=−52−412−1=−132−12=13m_{AD} = \frac{-\frac52-4}{\frac12-1} = \frac{-\frac{13}{2}}{-\frac12} = 13

y−4=13(x−1)   ⟹   y=13x−13+4=13x−9   ⟹   13x−y−9=0y-4 = 13(x-1) \ \implies\ y = 13x-13+4 = 13x-9 \ \implies\ 13x-y-9=0

  1. (ii) Altitude through A — perpendicular to BCBC, passing through A(1,4)A(1,4).

m⊥=−1mBC=−1−1/3=3m_\perp = -\frac{1}{m_{BC}} = -\frac{1}{-1/3} = 3

y−4=3(x−1)   ⟹   y=3x−3+4=3x+1   ⟹   3x−y+1=0y-4 = 3(x-1) \ \implies\ y = 3x-3+4 = 3x+1 \ \implies\ 3x-y+1=0

  1. (iii) Perpendicular bisector of BC — perpendicular to BCBC (same slope 33 as above), through midpoint D(12,−52)D\left(\tfrac12,-\tfrac52\right).

y−(−52)=3(x−12)   ⟹   y+52=3x−32y-\left(-\frac52\right) = 3\left(x-\frac12\right) \ \implies\ y+\frac52 = 3x-\frac32

y=3x−32−52=3x−4   ⟹   3x−y−4=0y = 3x-\frac32-\frac52 = 3x-4 \ \implies\ 3x-y-4=0

  1. Self-check. Median passes through both A(1,4)A(1,4): 13−4−9=013-4-9=0 ✓, and D(12,−52)D\left(\tfrac12,-\tfrac52\right): 13(12)−(−52)−9=6.5+2.5−9=013\left(\tfrac12\right)-\left(-\tfrac52\right)-9 = 6.5+2.5-9=0 ✓. Altitude passes through AA: 3−4+1=03-4+1=0 ✓, and mBC×3=−13×3=−1m_{BC}\times 3 = -\tfrac13\times3=-1 ✓ (perpendicular). Perpendicular bisector passes through DD: 3(0.5)−(−2.5)−4=1.5+2.5−4=03(0.5)-(-2.5)-4=1.5+2.5-4=0 ✓.
✓Final answer

(i) 13x−y−9=013x-y-9=0 (ii) 3x−y+1=03x-y+1=0 (iii) 3x−y−4=03x-y-4=0

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