Class 11: Cartesian System of Rectangular Co-ordinates

$\displaystyle \textbf{Question 1. }\text{The vertices of a triangle are }O(0,0),A(a,0)\text{ and }B(0,b).\text{ Write} \\ \text{the coordinates of its circumcentre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \triangle OAB\text{ is right-angled at }O$
$\displaystyle \text{Circumcentre is the mid-point of hypotenuse }AB$
$\displaystyle \therefore \text{Circumcentre}=\left(\frac{a+0}{2},\frac{0+b}{2}\right)$
$\displaystyle =\left(\frac{a}{2},\frac{b}{2}\right)$
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$\displaystyle \textbf{Question 2. }\text{In Q.No. 1, write the distance between the circumcentre and orthocentre} \\ \text{of }\triangle OAB.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }\triangle OAB\text{ is right-angled at }O,\text{ its orthocentre is }O(0,0)$
$\displaystyle \text{Circumcentre}=\left(\frac{a}{2},\frac{b}{2}\right)$
$\displaystyle \therefore \text{Distance}=\sqrt{\left(\frac{a}{2}\right)^2+\left(\frac{b}{2}\right)^2}$
$\displaystyle =\frac{1}{2}\sqrt{a^2+b^2}$
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$\displaystyle \textbf{Question 3. }\text{Write the coordinates of the orthocentre of the triangle formed by points } \\ (8,0),(4,6)\text{ and }(0,0).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(8,0),\ B(4,6),\ C(0,0)$
$\displaystyle \text{Slope of }BC=\frac{6-0}{4-0}=\frac{3}{2}$
$\displaystyle \therefore \text{Slope of altitude through }A=-\frac{2}{3}$
$\displaystyle \text{Equation of altitude through }A\text{ is}$
$\displaystyle y=-\frac{2}{3}(x-8)$
$\displaystyle \text{Slope of }AC=0$
$\displaystyle \therefore \text{Altitude through }B\text{ is vertical}$
$\displaystyle x=4$
$\displaystyle \text{Putting }x=4\text{ in }y=-\frac{2}{3}(x-8),$
$\displaystyle y=-\frac{2}{3}(4-8)=\frac{8}{3}$
$\displaystyle \therefore \text{Orthocentre}=\left(4,\frac{8}{3}\right)$
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$\displaystyle \textbf{Question 4. }\text{Three vertices of a parallelogram, taken in order, are }(-1,-6),(2,-5) \\ \text{ and }(7,2). \text{ Write the coordinates of its fourth vertex.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(-1,-6),B(2,-5),C(7,2)\text{ and }D(x,y)$
$\displaystyle \text{For a parallelogram, }D=A+C-B$
$\displaystyle D=(-1,-6)+(7,2)-(2,-5)$
$\displaystyle D=(4,1)$
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$\displaystyle \textbf{Question 5. }\text{If the points }(a,0),(at_1^2,2at_1)\text{ and }(at_2^2,2at_2)\text{ are collinear,} \\ \text{write the value of }t_1t_2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the three points are collinear, their area is }0$
$\displaystyle \therefore \begin{vmatrix}a&0&1\\ at_1^2&2at_1&1\\ at_2^2&2at_2&1\end{vmatrix}=0$
$\displaystyle a\begin{vmatrix}1&0&1\\ t_1^2&2t_1&1\\ t_2^2&2t_2&1\end{vmatrix}=0$
$\displaystyle \begin{vmatrix}1&0&1\\ t_1^2&2t_1&1\\ t_2^2&2t_2&1\end{vmatrix}=0$
$\displaystyle 2t_1-2t_2+2t_1^2t_2-2t_1t_2^2=0$
$\displaystyle 2(t_1-t_2)+2t_1t_2(t_1-t_2)=0$
$\displaystyle 2(t_1-t_2)(1+t_1t_2)=0$
$\displaystyle \therefore t_1t_2=-1$
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$\displaystyle \textbf{Question 6. }\text{If the coordinates of the mid-points of sides }AB\text{ and } AC\text{ of a } \\ \triangle ABC\text{ are }(3,5)\text{ and }(-3,-3)\text{ respectively, then write the length of side }BC.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the mid-points of }AB\text{ and }AC\text{ be }D(3,5)\text{ and }E(-3,-3)$
$\displaystyle DE\parallel BC\text{ and }DE=\frac{1}{2}BC$
$\displaystyle DE=\sqrt{(3+3)^2+(5+3)^2}$
$\displaystyle =\sqrt{6^2+8^2}=10$
$\displaystyle \therefore BC=2DE=20$
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$\displaystyle \textbf{Question 7. }\text{Write the coordinates of the circumcentre of a triangle whose centroid} \\ \text{and orthocentre are at }(3,3)\text{ and }(-3,5)\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Centroid }G\text{ divides the line joining orthocentre }H\text{ and circumcentre } \\ O\text{ in the ratio }2:1$
$\displaystyle G=\frac{H+2O}{3}$
$\displaystyle (3,3)=\frac{(-3,5)+2(x,y)}{3}$
$\displaystyle (9,9)=(-3,5)+(2x,2y)$
$\displaystyle 2x=12,\qquad 2y=4$
$\displaystyle \therefore O=(6,2)$
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$\displaystyle \textbf{Question 8. }\text{Write the coordinates of the incentre of the triangle having its vertices} \\ \text{at }(0,0),(5,0)\text{ and }(0,12).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(0,0),B(5,0),C(0,12)$
$\displaystyle BC=13,\quad CA=12,\quad AB=5$
$\displaystyle \text{Incentre}=\left(\frac{13\cdot0+12\cdot5+5\cdot0}{13+12+5},\frac{13\cdot0+12\cdot0+5\cdot12}{13+12+5}\right)$
$\displaystyle =\left(\frac{60}{30},\frac{60}{30}\right)$
$\displaystyle =(2,2)$
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$\displaystyle \textbf{Question 9. }\text{If the points }(1,-1),(2,-1)\text{ and }(4,-3)\text{ are the mid-points of the} \\ \text{sides of a triangle, then write the coordinates of its centroid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The centroid of a triangle is also the centroid of its medial triangle}$
$\displaystyle \therefore G=\left(\frac{1+2+4}{3},\frac{-1-1-3}{3}\right)$
$\displaystyle =\left(\frac{7}{3},-\frac{5}{3}\right)$
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$\displaystyle \textbf{Question 10. }\text{Write the area of the triangle having vertices at }(a,b+c), \\ (b,c+a),(c,a+b).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area}=\frac{1}{2}\left|\begin{vmatrix}a&b+c&1\\ b&c+a&1\\ c&a+b&1\end{vmatrix}\right|$
$\displaystyle =\frac{1}{2}|a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)|$
$\displaystyle =\frac{1}{2}|a(c-b)+b(a-c)+c(b-a)|$
$\displaystyle =\frac{1}{2}|0|$
$\displaystyle =0$
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