Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum
The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.
Advertisement Remove all ads
Solution
Let A(–1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.
∴ Coordiantes of the mid-point of AC = Coordinates of the mid-point of BD
`\Rightarrow ( \frac{-1+2}{2},\ \frac{0+2}{2})=(\frac{3+x}{2},\ \frac{1+y}{2})`
`\Rightarrow ( \frac{1}{2},\ 1)=( \frac{3+x}{2},\frac{y+1}{2})`
`\Rightarrow \frac{3+x}{2}=\frac{1}{2}\text{ and }\frac{y+1}{2}=1`
⇒ x = – 2 and y = 1
Hence, the fourth vertex of the parallelogram is (–2, 1).
Concept: Section Formula
Is there an error in this question or solution?