Advertisements
Advertisements
प्रश्न
Find the equation of a line whose slope and y-intercept are m = `(-6)/5`, c = 3
उत्तर
Equation of line with slope and y intercept
m = `(-6)/5`, c = 3
y = mx + c
y = `(-6)/5"x" + 3`
5y + 6x - 15 = 0
⇒ 6x + 5y = 15
APPEARS IN
संबंधित प्रश्न
In the figure given below, the line segment AB meets X-axis at A and Y-axis at B. The point P(-3, 4) on AB divides it in the ratio 2:3. Find the coordinates of A and B.
Find, which of the following points lie on the line x – 2y + 5 = 0 :
(1, 3)
The line through P (5, 3) intersects y-axis at Q.
Find the co-ordinates of Q.
A(1, 4), B(3, 2) and C(7, 5) are vertices of a triangle ABC. Find the co-ordinates of the centroid of triangle ABC.
A(7, −1), B(4, 1) and C(−3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP : CP = 2 : 3.
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.
Find the value of a if the line 4 x = 11 - 3y passes through the point (a, 5)
X(4,9), Y(-5,4) and Z(7,-4) are the vertices of a triangle. Find the equation of the altitude of the triangle through X.
ABCD is a square. The cooordinates of B and D are (-3, 7) and (5, -1) respectively. Find the equation of AC.
Find the equation of the straight line which has Y-intercept equal to 4/3 and is perpendicular to 3x – 4y + 11 = 0.