Question

The equation of line equally inclined to coordinate and passing through is (– 3, 2, – 5) is
$A.\dfrac{{x + 3}}{1} = \dfrac{{y - 2}}{1} = \dfrac{{z + 5}}{1}$
$B.\dfrac{{x + 3}}{1} = \dfrac{{y - 2}}{1} = \dfrac{{z + 5}}{{ - 1}}$
$C.\dfrac{{x + 3}}{{ - 1}} = \dfrac{{y - 2}}{1} = \dfrac{{z + 5}}{1}$
$D.\dfrac{{x + 3}}{{ - 1}} = \dfrac{{y - 2}}{1} = \dfrac{{z + 5}}{{ - 1}}$

Answer 1

Hint: We need to use the equation of line passing through a point is
$\dfrac{{x + {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z + {z_1}}}{n}$ 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
$\dfrac{{x + {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z + {z_1}}}{n}$
after putting the value of $(x_1, y_1, z_1)$ and (l, m, n) we have
$\dfrac{{x + 3}}{{ - 1}} = \dfrac{{y - 2}}{1} = \dfrac{{z + 5}}{{ - 1}}$
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.