find the equation of the straight line through the point of intersection of the lines 2y = 3 3 and use-y + 7 =0 and which & Parallel to 5x+4y = 13.
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Let the equation of the line having equal intercepts on the axes be
ax+ay=1
x+y=a....(1)
On solving equation 4x+7y−3=0 and 2x−3y+1=0 we obtain x=131andy=135
∴{131,135} is the point of intersection of the two given lines.
Since equation (1) passes through point {131,135}
{131+135}=a
⇒a=136
∴ Equation (1) becomes x+y=136,i.e.,13x+13y=6
Thus the required equation of the line is 13x+13y=6
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