Find the equation of the line passing through the centroid of traingle:PQR, with vertices P(3, 3), Q(-2, -6) and R(5, -3) and parallel to the line QR.
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We have to find the equation of the line passing through the centroid of triangle PQR with vertices P(3, 3) , Q(-2, -6) and R(5, -3) and the parallel to the line QR.
solution : centroid = [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3]
= [(3 - 2 + 5)/3, (3 - 6 - 3)/3]
= (2, -2)
slope of line QR = (y₃ - y₂)/(x₃ - x₂)
= (-3 + 6)/(5 +2)
= 3/7
because line passing through centroid is parallel to line QR
so slope of line passing through centroid = slope of line QR
now equation of line is ..
y - (-2) = 3/7(x - 2)
⇒7(y + 2) = 3(x - 2)
⇒7y + 14 = 3x - 6
⇒3x - 7y - 20 = 0
Therefore the equation of line passing through the centroid of triangle PQR is 3x - 7y - 20 = 0.
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