The coordinates of the vertices of triangle ABC are A(-2, 1), B(4, -2) and C(1, 8) respectively. If D(x, y) is the foot perpendicular from A to BC, find (a)...
Question 1 Report
The coordinates of the vertices of triangle ABC are A(-2, 1), B(4, -2) and C(1, 8) respectively. If D(x, y) is the foot perpendicular from A to BC, find
(a) an equation connecting x and y ;
(b) the unit vector in the direction of BC.
Answer Details
\(A(-2,1),\ B(4,-2),\ C(1,8)\). \(D(x,y)\) is the foot of the perpendicular from \(A\) to \(BC\).
(a) Direction of \(BC\): \(\vec{BC} = C - B = (1 - 4,\ 8 - (-2)) = (-3,\ 10)\). Vector \(\vec{AD} = (x + 2,\ y - 1)\). Since \(AD \perp BC\), their dot product is zero:
\[(x + 2)(-3) + (y - 1)(10) = 0\] \[-3x - 6 + 10y - 10 = 0 \;\Rightarrow\; 3x - 10y + 16 = 0\]The equation connecting \(x\) and \(y\) is \(3x - 10y + 16 = 0\).
(b) \(|\vec{BC}| = \sqrt{(-3)^2 + 10^2} = \sqrt{109}\). Unit vector in the direction of \(BC\):
\[\hat{BC} = \frac{1}{\sqrt{109}}\begin{pmatrix} -3 \\ 10 \end{pmatrix} = \begin{pmatrix} -\tfrac{3}{\sqrt{109}} \\[4pt] \tfrac{10}{\sqrt{109}} \end{pmatrix}\]Download The App On Google Playstore
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