The line (x/a)-(y/b)=1 cuts the x- axis at P . The equation of the line through P perpendicular to the given line is:

Question

The line (x/a)-(y/b)=1 cuts the x- axis at P . The equation of the line through P perpendicular to the given line is: (A)  x+y=ab  (B)  x+y=a+b  (C)  ax+by=a2  (D)  bx+ay=b2

Tardigrade Admin · Accepted Answer

Given equation of a line is Its slope is (b/a). ∴ The equation of line passing through a point P(a, 0) and perpendicular to the given line whose slope is -(a/b), is y-0=-(a/b)(x-a) b y+a x=a2

Q. The line $\frac{x}{a} - \frac{y}{b} = 1$ cuts the $x -$ axis at $P$ . The equation of the line through $P$ perpendicular to the given line is:

$x + y = ab$

$x + y = a + b$

$a x + b y = a^{2}$

$b x + a y = b^{2}$

Given equation of a line is
Its slope is $\frac{b}{a}$ .
$∴$ The equation of line passing through a point
$P (a, 0)$ and perpendicular to the given line whose slope is
$- \frac{a}{b}$ , is $y - 0 = - \frac{a}{b} (x - a)$
$b y + a x = a^{2}$

Q. The line ax​−by​=1 cuts the x− axis at P . The equation of the line through P perpendicular to the given line is:

x+y=ab

x+y=a+b

ax+by=a2

bx+ay=b2