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The x-coordinate of a point P is twice itsy-coordinate. If P is equidistant from Q(2, -5)and R(-3, 6), find the co-ordinates of P. |
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Answer» it's simple dude...... let the ordinates of P be y the then abcissa will be 2y {(2y-2)^2+(y+5)}^1/2={(2y+3)^2+(y-6)^2}^1/2 4y^2+4-8y+y^2+25+10y=4y^2+9+12y+y^2+36-12y 4-8y+25+10y=9+12y+36-12y 2y-6y+=45-29 4y=16 y=4 hence coordinates are(8,4) {EASIEST question} |
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