The distance between $$1$$,$$1$$ and the point of intersection of two lines $$2x+3y+c=0$$ and $$3x+2y+c=0$$ is less than $$22then$$

Question

The distance between $$1$$,$$1$$ and the point of intersection of two lines $$2x+3y+c=0$$ and  $$3x+2y+c=0$$ is less than $$22then$$

Infinity Learn · Accepted Answer

To get the point of intersection of two lines, eliminate  $$x32x+3y+c$$−$$23x+2y+c=06x+9y+3c$$−$$6x$$−$$4y$$−$$2c=05y+c=0y=$$−$$c5$$ Substitute $$y=$$−$$c5$$ in the equation $$2x+3y+c=02x+3$$−$$c5+c=010x$$−$$3c+5c=010x+2c=0x=$$−$$c5$$ Hence, the point of intersection is −$$c5$$,−$$c5$$ Given that the distance between (1$$,$$1) and −$$c5$$,−$$c5$$ is $$1+c52+1+c52$$<$$22It$$ implies that $$21+c52$$<$$22$$  Squaring and then simplify   $$21+c52$$<$$81+c52$$<$$4Apply$$ square root on both sides −$$2$$<$$1+c5$$<$$2$$−$$3$$

The distance between $(1, 1)$ and the point of intersection of two lines $2 x + 3 y + c = 0$ and $3 x + 2 y + c = 0$ is less than $2 \sqrt{2}$ then

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The distance between 1,1 and the point of intersection of two lines 2x+3y+c=0 and 3x+2y+c=0 is less than 22then

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The distance between $(1, 1)$ and the point of intersection of two lines $2 x + 3 y + c = 0$ and $3 x + 2 y + c = 0$ is less than $2 \sqrt{2}$ then