Question
Question: The equation of line equally inclined to coordinate and passing through is (– 3, 2, – 5) is \(A.\d...
The equation of line equally inclined to coordinate and passing through is (– 3, 2, – 5) is
A.1x+3=1y−2=1z+5
B.1x+3=1y−2=−1z+5
C.−1x+3=1y−2=1z+5
D.−1x+3=1y−2=−1z+5
Solution
We need to use the equation of line passing through a point is
lx+x1=my−y1=nz+z1 and before this we need to calculate the inclination is l, m, n. The general equation of a straight line is y = mx + c, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis.
Complete step-by-step answer:
It is given a line equally inclined to the coordinate axis and also passing through (– 3, 2 – 5).
So,
Let (x, y, z) = (– 3, 2, – 5) respectively.
It is also given that the line is equally inclined.
l = –1, m = 1 and n = – 1
We know that, the equation of line passing through (x, y, z) and having direction cosines l, m, n is
lx+x1=my−y1=nz+z1
after putting the value of (x1,y1,z1) and (l, m, n) we have
−1x+3=1y−2=−1z+5
Hence the answer is option D
Note: In this type of question, we need to know the equation for a line passing through a point. We can find the equation of a straight line when given the gradient and a point on the line by using the formula:
y−b=m (x−a)
Where m is the gradient and (a, b) is on the line.