Question

12. Find the x and y intercepts of the straight line
(i) 5x=3y-15 = 0 (ii) 2x -y + 16 = 0 (iii) 3x +10y + 4 = 0

Answer 1

Solution:-

given by:-(ii) 2x -y + 16 = 0
$(2) \: \: 2x - y + 16 = 0\\ 2x - y = - 16 \\ - \frac{2x}{16} + \frac{y}{16} = 1 \\ - \frac{x}{8} + \frac{y}{16} = 1 \\ by \: intercepts \: form \: of \: stright \: line \\ \frac{x}{a} + \frac{y}{b} = 1 \\ here \\ \\ a = - 8 \\ b = 16$
(a and b the intercepts x axis and y axis)ans


given by (iii) 3x +10y + 4 = 0
$3x + 10y + 4 = 0 \\ 3x + 10y = - 4 \\ - \frac{3x}{4} - \frac{10y}{4} = 1 \\ \frac{ - 3x}{4} - \frac{5y}{2} = 1 \\ by \: intercepts \: form \: \\ \\ \frac{x}{a} + \frac{y}{a} = 1 \\ \\ a = \frac{ - 3}{4} \\ b = \frac{ - 5}{2}$

(here a and b the intercepts) ans



☆i hope its help☆

Answer 2

Solution :

*******************"********************

Equation of a line whose ,

x-intercept = a ,

y-intercept = b , is

x/a + y/b = 1

****************************************

i ) Convert 5x = 3y - 15 into

intercept form ,

5x - 3y = - 15

divide each term with -15, we get

5x/(-15) + (-3y)/(-15) = (-15)/(-15)

=> x/(-3) + y/5 = 1

x-intercept = a = -3 ,

y-intercept = b = 5 ,

ii ) Given 2x - y + 16 = 0

=> 2x - y = -16

divide each term with -16 , we get

=>2x/(-16) + (-y)/(-16) = (-16)/(-16)

=> x/(-8) + y/(16) = 1

Therefore ,

x-intercept = a = -8 ,

y-intercept = b = 16 ,

iii ) Given 3x + 10y + 4 =0

=> 3x + 10y = -4

divide each term with -4 , we get

=> 3x/(-4) + 10y/(-4) = (-4)/(-4)

=> x/(-4/3) + y/(-2/5) = 1

x-intercept = a = -4/3 ,

y-intercept = b = -2/5

••••