Question

if a line cuts a intercepts +2 and 3 on x-axis and y-axis then prove that its equation is 3x-2y+6=0 find out whether the point are (4,9) lies on this line.​

Answer 1

$\bf\large{\underline{\underline{Question:-}}}$if a line cuts a intercepts +2 and 3 on x-axis and y-axis then prove that its equation is 3x-2y+6=0 find out whether the point are (4,9) lies on this line.

$\bf\large{\underline{\underline{Solution:-}}}$

• Here , a= -2
• B= 3

The equation of line is

$\tt→ {x}{-2}+\frac{y}{3}=1\\\tt→ \frac {x(-6)}{2}+\frac{y×(-6)}{2}=1×(-6)\\\tt→ 3x-3y,=-6\\\tt→ 3x-2y+6=0.$

If,

the point (4,9) lies on the line,

then x=4 and y=9 must have satisfy the equation of the line.

Here , in this equation we have

L.H.S = 3x-2y+6=3×4-2×9+6

12-18+6

=0 R.H.S

PROVED

Answer 2

Step-by-step explanation:

$3x - 2 = 0 \\ y = 0 \: \: \: \: \: 3x = - 6 = > x = - 2 \\ x = 0 \: \: \: \: \: - 2y = - 6 = > y = 3 \\ ab = \sqrt{ {(0 + 2)}^{2} + {(3 - 0)}^{2} } \\ = > \sqrt{4 + 9} \\ = > \sqrt{13} \\ equation = > {(x + 2)}^{2} + {(y - 0)}^{2} \\ = > {( \sqrt{13)} }^{2}$

${x}^{2} + 4 + 4x + {y}^{2} = > 13 \\ {x}^{2} + {y}^{2} + 4x - 9 = > 0$

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