Question 1011755
The side JT is perpendicular to the
side JK.
Find the slope of JK:
J(-2,-3) and K(3,-3)
{{{ m[1] = ( -3 -(-3) ) / ( 3 -(-2) ) }}}
{{{ m[1] = 0 }}}
Zero slope is a horizontal line, so any
line perpendicular to it is a vertical line
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The side JT goes through J( -2,-3 ), so
it's equation is {{{ x = -2 }}}
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The length of horizontal line JK is:
{{{ d = sqrt( ( 3 -(-2))^2 + ( -3 -(-3) )^2 ) }}}
{{{ d = sqrt( 5^2 + 0^2 ) }}}
{{{ d = sqrt( 25 ) }}}
{{{ d = 5 }}}
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Since it's a 45-45-90 triangle, side JT also
is {{{ 5 }}} units long.
It's a vertical line and the x-coordinate is {{{ x = -2 }}}
Point T is:
T( -2, 5 )
Note this is in quadrant 2 also