Coordinate Geometry
Category : 10th Class
Coordinate Geometry
Coordinate Geometry
\[PQ\,=\,\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}\,+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
\[\frac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\,\frac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}}\]
9. Area of a Triangle: The area of a \[\Delta \,PQR\] having vertices points \[P({{x}_{1}},\,\,{{y}_{1}}),\,Q({{x}_{2}},\,\,{{y}_{2}})\] and \[R({{x}_{3}},\,\,{{y}_{3}})\] is the numerical value of the expression \[\left| \frac{1}{2}\{{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})\} \right|\]
Snap Test
(a) \[\left( 0,\,\,\frac{21}{5} \right)\]
(b) \[\left( 1,\,\,\frac{21}{5} \right)\]
(c) \[\left( 1,\,\,\frac{21}{4} \right)\]
(d) \[\left( 2,\,\,\frac{21}{4} \right)\]
(e) None of these
Ans. (a)
Explanation: Let AB be the line segment with end points \[A\left( 6,\text{ }3 \right)\text{ }and\text{ }B\text{ }\left( -\text{ }4,\text{ }5 \right).\]
Then, \[\left( {{x}_{1}}=\text{ }6,\text{ }{{y}_{1}}=\text{ }3 \right)\] and \[\left( {{x}_{2}}=-\text{ }4,\text{ }{{y}_{2}}=\text{ }5 \right)\].
Also, \[{{m}_{1}}=\text{ }3\text{ }and\text{ }{{m}_{2}}=\text{ }2\].
Let the required point be \[P\text{ }\left( x,\text{ }y \right)\].
By section formula:
\[x=\frac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\,\,y\,\,=\frac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}}\]
\[x\,\,=\,\,\frac{3\times (-4)+2\times 6}{(3+2)},\,\,y\,\frac{(3\times 5+2\times 3)}{(3+2)}\,\,\,\Rightarrow \,x=0,\,\,y=\frac{21}{5}\]
Hence, the required point is \[P\left( 0,\,\,\frac{21}{5} \right)\]
(a) \[3\text{ }:\text{ }1\]
(b) \[1\text{ }:\text{ }1\]
(c) \[2\text{ }:\text{ }1\]
(d) \[2\text{ }:\text{ }2\]
(e) None of these
Ans. (b)
Explanation: Let P \[\left( -\text{ }3,\text{ }k \right)\] divide the line segment \[A\left( -\text{ }5,\text{ }-4 \right)\] and \[B\text{ }\left( -\text{ }2,\text{ }3 \right)\] in the ratio \[m\text{ }:\text{ }1\],
Then the coordinates of P are given by \[\left( \frac{-2m-5}{m+1},\,\,\frac{3m-4}{m+1} \right)\].
But, we are given \[P\left( -\text{ }3,\text{ }k \right)\].
\[\therefore \,\,\,\,\,\frac{-2m-5}{m+1}=\,\,-3\,\,and\,\,\,\frac{3m-4}{m+1}\,\,=\,\,k\]
Now, \[\frac{-2m-5}{m+1}=-\text{ }3\text{ }\Rightarrow \,\,-2m\text{ }-\text{ }5\text{ }=-\text{ }3m\text{ }-\text{ }1\text{ }\Rightarrow \text{ }2m\text{ }=\text{ }4\text{ }\Rightarrow \text{ }m=2.\]
So, the required ratio is \[2\text{ }:\text{ }1\].
(a) \[\left( 1,\text{ }-1 \right),\text{ }\left( 5,\text{ }2 \right)\text{ }and\text{ }\left( 3,\text{ }5 \right)\]
(b) \[\left( 1,\,-1 \right),\text{ }\left( 5,\text{ }2 \right)\text{ }and\text{ }\left( 5,\text{ }5 \right)\]
(c) \[\left( 1,\text{ }-1 \right),\text{ }\left( 5,\text{ }2 \right)\text{ }and\text{ }\left( 9,\text{ }5 \right)\]
(d) \[\left( 1,-\text{ }1 \right),\left( 5,\text{ }2 \right)\text{ }and\text{ }\left( 6,\text{ }5 \right)\]
(e) None of these
Ans. (c)
Explanation: Let \[P\left( 1,\text{ }-\text{ }1 \right),\text{ }Q\left( 5,\text{ }2 \right)\text{ }and\text{ }R\left( 9,\text{ }5 \right)\] be the given points. Then,
\[PQ=\sqrt{{{(5-1)}^{2}}+{{[2-(-1)]}^{2}}}\,\,=\sqrt{{{4}^{2}}+{{3}^{2}}}\,\,=\,\,\sqrt{25}=5\,units,\]
\[QR=\sqrt{{{(9-5)}^{2}}+(5-2){{]}^{2}}}\,\,=\,\,\sqrt{{{4}^{2}}+{{3}^{2}}}\,\,=\,\,\sqrt{25}=5\,units,\]
\[PR\,\,\,\,=\,\,\,\,\sqrt{{{(9-1)}^{2}}+{{[5-(-1)]}^{2}}}\,\,=\,\,\,\sqrt{{{8}^{2}}+{{6}^{2}}}\,\,=\,\,\,\sqrt{100}\,\,\,=\,\,10\,\,units\]
\[\therefore \,\,\,\,\,PQ+QR\text{ }=\left( 5\text{ }+\text{ }5 \right)\text{ }units\text{ }=10\text{ }units\text{ }=\text{ }PR\]
Hence, the given points A, B, C are collinear.
(a) \[a\text{ }=\text{ }1,\text{ }b\text{ }=\text{ }3\]
(b) \[a\text{ }=\text{ }1,\text{ }b\text{ }=\text{ }5\]
(c) \[a\text{ }=\text{ }3,\text{ }b\text{ }=\text{ }1\]
(d) \[a\text{ }=\text{ }2,\text{ }b\text{ }=\text{ }4\]
(e) None of these
Ans. (a)
Explanation: In the following figure. Let \[P\left( -2,\text{ }-1 \right),\text{ }Q\text{ }\left( a,\text{ }0 \right),\text{ }R\left( 4,\text{ }b \right)\text{ }and\text{ }S\left( 1,\text{ }2 \right)\] be the vertices of a parallelogram PQRS.
Join the diagonals PR and QS, intersecting each other at the point O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, 0 is the mid-point of AC as well as that of BD.
Now, mid-point of PR is \[\left( \frac{-2+4}{2},\,\,\frac{-1+b}{2} \right)\,i.e.\,\left( 1,\,\,\frac{b-1}{2} \right)\]
And, mid-point of QS is \[\left( \frac{a+1}{2},\,\,\frac{0+2}{2} \right)\,i.e.\,\left( \frac{a+1}{2}\,\,,\,\,1\, \right)\].
\[\therefore \,\,\,\,\frac{a+1}{2}=1\,\,\,and\,\,\,\frac{b-1}{2}\,\,\,=\,\,\Rightarrow \,\,a=1\,\,and\,\,b=3.\]
(a) 3 units, 3 units (b) 6 units, 6 units
(c) 5 units, 5 units (d) 4 units, 4 units
(e) None of these
Ans. (c)
Explanation: in the following figure, Let \[A\left( -1,\text{ }3 \right),\text{ }B\left( 1,\text{ }-1 \right)\text{ }and\text{ }C\left( 5,\text{ }1 \right)\] be the three vertices of a \[\Delta \,ABC\] Let D and E be the mid-points of the sides BC and AB respectively.
The coordinates of \[D\,\,=\,\,\left( \frac{1+5}{2},\,\,\frac{-1+1}{2} \right)\]
i.e. \[\left( 3,\text{ }0 \right)\] and those of E are \[\left( \frac{-1+1}{2},\,\,\frac{3-1}{2} \right)\] i.e. \[\left( 0,\text{ }1 \right)\].
\[\therefore \] Length of median through \[A\left( -1,\text{ }3 \right)\]
\[=\,\,\,AD\,\,\,=\sqrt{{{[3-(-1)]}^{2}}+{{(0-3)}^{2}}}=\sqrt{{{4}^{2}}+{{(-3)}^{2}}}\,\,=\,\,\,\sqrt{16+9}\,\,\,=\,\,\,\sqrt{25}\,\,=\,\,5\,\,units.\]
Length of median through \[C\text{ }\left( 5,\text{ }1 \right)\]
\[=\,\,CE\,\,=\,\,\,\sqrt{{{(0-5)}^{2}}+{{(1+1)}^{2}}}\,\,\,=\,\,\,\sqrt{{{(-5)}^{2}}+0}\,\,\,\,=\,\,\sqrt{25}\,\,\,=\,\,5\,\,units\].
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