Advertisements
Advertisements
Question
ABCD is a square. The cooordinates of B and D are (-3, 7) and (5, -1) respectively. Find the equation of AC.
Solution
Slope of BD = `(7 + 1)/(-3 -5)` = 1
Slope of AC = 1
Mid point of AC = mid point of BD
O(x,y) = `((5 - 3)/2 , (-1 + 7)/2)` = (1,3)
Equation of AC ⇒ `("y" - "y"_1)/("x" - "x"_1)` = slope
⇒ `("y" - 3)/("x" - 1) = 1`
⇒ x - 1 = y - 3
⇒ x - y + 2 = 0
⇒ y = x + 2
APPEARS IN
RELATED QUESTIONS
Find, if point (-2,-1.5) lie on the line x – 2y + 5 = 0
State, true or false :
The point (8, 7) lies on the line y – 7 = 0
The line given by the equation `2x - y/3 = 7` passes through the point (k, 6); calculate the value of k.
Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
The line through P (5, 3) intersects y-axis at Q.
Find the co-ordinates of Q.
The vertices of a triangle ABC are A(0, 5), B(−1, −2) and C(11, 7). Write down the equation of BC. Find:
- the equation of line through A and perpendicular to BC.
- the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.
The line segment formed by two points A (2,3) and B (5, 6) is divided by a point in the ratio 1 : 2. Find, whether the point of intersection lies on the line 3x - 4y + 5 = 0.
Find the inclination of a line whose gradient is 3.0777
ABCD is rhombus. The coordinates of A and C ae (3,7) and (9,15). Find the equation of BD.
Find the value of a for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.