88279
If the vertices of a triangle are \(A(0,4,1), B(2,3\), \(-1)\) and \(C(4,5,0)\), then the orthocentre of \(\triangle \mathrm{ABC}\), is
1 \((4,5,0)\)
2 \((2,3,-1)\)
3 \((-2,3,-1)\)
4 \((2,0,2)\)
Explanation:
(B) : Vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(0,4,1), \mathrm{B}(2,3,-1)\) and \(\mathrm{C}(4,5,0)\). \(\mathrm{AB}=\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}}\) \(\mathrm{AB}=\sqrt{4+1+4}\) \(\mathrm{AB}=\sqrt{9}\) \(\mathrm{AB}=3\) \(\mathrm{BC}=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}}\) \(\mathrm{BC}=\sqrt{4+4+1}\) \(\mathrm{BC}=\sqrt{9}\) \(\mathrm{BC}=3\) \(\mathrm{CA}=\sqrt{(4-0)^{2}+(5-4)^{2}+(0-1)^{2}}\) \(\mathrm{CA}=\sqrt{16+1+1}\) \(\mathrm{CA}=3 \sqrt{2}\) \(\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}\) \(\therefore \triangle \mathrm{ABC}\) is a right angled triangle. We know that, the orthocenter of a right angled triangle is the vertex containing the right angle. \(\therefore\) Orthocenter is point B \((2,3,-1)\).
VITEEE-2014
Co-Ordinate system
88280
The number of integral points (integral point means both the coordinates should be integer exactly in the interior of the triangle with vertices \((0,0),(0,21)\) and \((21,0)\) is
1 133
2 190
3 233
4 105
Explanation:
(B) : Let the vertices of triangle be \(\mathrm{A}(21,0), \mathrm{B}(0,21)\) and \(\mathrm{O}(0,0)\) Thus, any point in the interior of the triangle lies in first quadrant. \(\therefore \quad a>0 \& b>0\) Point \((a, b)\) lies on the same side of the \(A B\) where lies. For \((0,0)\) therefore \(x+y \leq 21\) \(a+b-21\lt 0\) \(\Rightarrow \quad \mathrm{a}+\mathrm{b}\lt 21\) For \(\quad \mathrm{a}=1 ; \mathrm{b}\lt 21-1 \Rightarrow \mathrm{b}\lt 20\) \(b \in[1,19]\) total 19 integral value. For \(\quad a=2 ; b\lt 21-2 \Rightarrow b\lt 19\) Similarly \(b \in[1,18]\) total 18 integral values for \(\quad a=19 ; b\lt 21-19 \Rightarrow b\lt 2\) \(\mathrm{b}=1 \quad 1\) integral values Thus Number of integral points \(=19+18+\ldots .+1\) \(=\frac{19(19+1)}{2}=\frac{19 \times 20}{2}=190\) Thus there are total 190 integral points. Which lies inside the triangle.
VITEEE-2013
Co-Ordinate system
88281
In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio
1 \(2: 3: 5\)
2 \(1: 2: 3\)
3 \(1: 3: 7\)
4 \(3: 7: 9\)
Explanation:
(B) : We have, area of \(\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^{2}, \mathrm{~s}=\frac{3 \mathrm{a}}{2}\) In radius \(\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{a}}{2 \sqrt{3}}\) Circumradius \(\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^{3}}{\sqrt{3} \mathrm{a}^{2}}=\frac{\mathrm{a}}{\sqrt{3}}\) and exradius \(r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3} \mathrm{a}^{2} / 4}{\mathrm{a} / 2}\) \(=\frac{\sqrt{3}}{2} \mathrm{a}\) \(\therefore\) Required ratio \(=\mathrm{r}: \mathrm{R}: \mathrm{r}_{1}\) \(=\frac{\mathrm{a}}{2 \sqrt{3}}: \frac{\mathrm{a}}{\sqrt{3}}: \frac{\sqrt{3}}{2} \mathrm{a}=1: 2: 3\).
VITEEE-2011
Co-Ordinate system
88282
Let \(O=(0,0), A=(a, 11)\) and \(B=(b, 37)\) are the vertices of an equilateral triangle \(O A B\), then and batisfy the relation
1 \(\left(a^{2}+b^{2}\right)-4 a b=138\)
2 \(\left(a^{2}+b^{2}\right)-a b=124\)
3 \(\left(a^{2}+b^{2}\right)+3 a b=130\)
4 \(\left(a^{2}+b^{2}\right)-3 a b=138\)
Explanation:
(A) : Given, co-ordinate of vertices of equilateral triangle \(\mathrm{O}=(0,0), \mathrm{A}=(\mathrm{a}, 11), \mathrm{B}=(\mathrm{b}, 37)\) Let \(\mathrm{c}\) is the middle point of side \(\mathrm{AB}\) \(\therefore \quad \mathrm{C}=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, 24\right)\) \(\therefore \quad O C=\sqrt{3} C A\) \((\mathrm{OC})^{2}=(\sqrt{3}(\mathrm{CA}))^{2} \quad \text { [squaring both side] }\) \((\mathrm{OC})^{2}=3(\mathrm{CA})^{2}\) \(\left(\frac{\mathrm{a}+\mathrm{b}}{2}-0\right)^{2}+(24-0)^{2}=3\left[\left(\mathrm{a}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)^{2}+(11-24)^{2}\right]\) \(\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{4}+576=3\left[\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)^{2}+(-13)^{2}\right]\) \(\mathrm{a}^{2}+\mathrm{b}^{2}-4 \mathrm{ab}=138\)
88279
If the vertices of a triangle are \(A(0,4,1), B(2,3\), \(-1)\) and \(C(4,5,0)\), then the orthocentre of \(\triangle \mathrm{ABC}\), is
1 \((4,5,0)\)
2 \((2,3,-1)\)
3 \((-2,3,-1)\)
4 \((2,0,2)\)
Explanation:
(B) : Vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(0,4,1), \mathrm{B}(2,3,-1)\) and \(\mathrm{C}(4,5,0)\). \(\mathrm{AB}=\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}}\) \(\mathrm{AB}=\sqrt{4+1+4}\) \(\mathrm{AB}=\sqrt{9}\) \(\mathrm{AB}=3\) \(\mathrm{BC}=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}}\) \(\mathrm{BC}=\sqrt{4+4+1}\) \(\mathrm{BC}=\sqrt{9}\) \(\mathrm{BC}=3\) \(\mathrm{CA}=\sqrt{(4-0)^{2}+(5-4)^{2}+(0-1)^{2}}\) \(\mathrm{CA}=\sqrt{16+1+1}\) \(\mathrm{CA}=3 \sqrt{2}\) \(\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}\) \(\therefore \triangle \mathrm{ABC}\) is a right angled triangle. We know that, the orthocenter of a right angled triangle is the vertex containing the right angle. \(\therefore\) Orthocenter is point B \((2,3,-1)\).
VITEEE-2014
Co-Ordinate system
88280
The number of integral points (integral point means both the coordinates should be integer exactly in the interior of the triangle with vertices \((0,0),(0,21)\) and \((21,0)\) is
1 133
2 190
3 233
4 105
Explanation:
(B) : Let the vertices of triangle be \(\mathrm{A}(21,0), \mathrm{B}(0,21)\) and \(\mathrm{O}(0,0)\) Thus, any point in the interior of the triangle lies in first quadrant. \(\therefore \quad a>0 \& b>0\) Point \((a, b)\) lies on the same side of the \(A B\) where lies. For \((0,0)\) therefore \(x+y \leq 21\) \(a+b-21\lt 0\) \(\Rightarrow \quad \mathrm{a}+\mathrm{b}\lt 21\) For \(\quad \mathrm{a}=1 ; \mathrm{b}\lt 21-1 \Rightarrow \mathrm{b}\lt 20\) \(b \in[1,19]\) total 19 integral value. For \(\quad a=2 ; b\lt 21-2 \Rightarrow b\lt 19\) Similarly \(b \in[1,18]\) total 18 integral values for \(\quad a=19 ; b\lt 21-19 \Rightarrow b\lt 2\) \(\mathrm{b}=1 \quad 1\) integral values Thus Number of integral points \(=19+18+\ldots .+1\) \(=\frac{19(19+1)}{2}=\frac{19 \times 20}{2}=190\) Thus there are total 190 integral points. Which lies inside the triangle.
VITEEE-2013
Co-Ordinate system
88281
In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio
1 \(2: 3: 5\)
2 \(1: 2: 3\)
3 \(1: 3: 7\)
4 \(3: 7: 9\)
Explanation:
(B) : We have, area of \(\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^{2}, \mathrm{~s}=\frac{3 \mathrm{a}}{2}\) In radius \(\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{a}}{2 \sqrt{3}}\) Circumradius \(\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^{3}}{\sqrt{3} \mathrm{a}^{2}}=\frac{\mathrm{a}}{\sqrt{3}}\) and exradius \(r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3} \mathrm{a}^{2} / 4}{\mathrm{a} / 2}\) \(=\frac{\sqrt{3}}{2} \mathrm{a}\) \(\therefore\) Required ratio \(=\mathrm{r}: \mathrm{R}: \mathrm{r}_{1}\) \(=\frac{\mathrm{a}}{2 \sqrt{3}}: \frac{\mathrm{a}}{\sqrt{3}}: \frac{\sqrt{3}}{2} \mathrm{a}=1: 2: 3\).
VITEEE-2011
Co-Ordinate system
88282
Let \(O=(0,0), A=(a, 11)\) and \(B=(b, 37)\) are the vertices of an equilateral triangle \(O A B\), then and batisfy the relation
1 \(\left(a^{2}+b^{2}\right)-4 a b=138\)
2 \(\left(a^{2}+b^{2}\right)-a b=124\)
3 \(\left(a^{2}+b^{2}\right)+3 a b=130\)
4 \(\left(a^{2}+b^{2}\right)-3 a b=138\)
Explanation:
(A) : Given, co-ordinate of vertices of equilateral triangle \(\mathrm{O}=(0,0), \mathrm{A}=(\mathrm{a}, 11), \mathrm{B}=(\mathrm{b}, 37)\) Let \(\mathrm{c}\) is the middle point of side \(\mathrm{AB}\) \(\therefore \quad \mathrm{C}=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, 24\right)\) \(\therefore \quad O C=\sqrt{3} C A\) \((\mathrm{OC})^{2}=(\sqrt{3}(\mathrm{CA}))^{2} \quad \text { [squaring both side] }\) \((\mathrm{OC})^{2}=3(\mathrm{CA})^{2}\) \(\left(\frac{\mathrm{a}+\mathrm{b}}{2}-0\right)^{2}+(24-0)^{2}=3\left[\left(\mathrm{a}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)^{2}+(11-24)^{2}\right]\) \(\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{4}+576=3\left[\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)^{2}+(-13)^{2}\right]\) \(\mathrm{a}^{2}+\mathrm{b}^{2}-4 \mathrm{ab}=138\)
88279
If the vertices of a triangle are \(A(0,4,1), B(2,3\), \(-1)\) and \(C(4,5,0)\), then the orthocentre of \(\triangle \mathrm{ABC}\), is
1 \((4,5,0)\)
2 \((2,3,-1)\)
3 \((-2,3,-1)\)
4 \((2,0,2)\)
Explanation:
(B) : Vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(0,4,1), \mathrm{B}(2,3,-1)\) and \(\mathrm{C}(4,5,0)\). \(\mathrm{AB}=\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}}\) \(\mathrm{AB}=\sqrt{4+1+4}\) \(\mathrm{AB}=\sqrt{9}\) \(\mathrm{AB}=3\) \(\mathrm{BC}=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}}\) \(\mathrm{BC}=\sqrt{4+4+1}\) \(\mathrm{BC}=\sqrt{9}\) \(\mathrm{BC}=3\) \(\mathrm{CA}=\sqrt{(4-0)^{2}+(5-4)^{2}+(0-1)^{2}}\) \(\mathrm{CA}=\sqrt{16+1+1}\) \(\mathrm{CA}=3 \sqrt{2}\) \(\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}\) \(\therefore \triangle \mathrm{ABC}\) is a right angled triangle. We know that, the orthocenter of a right angled triangle is the vertex containing the right angle. \(\therefore\) Orthocenter is point B \((2,3,-1)\).
VITEEE-2014
Co-Ordinate system
88280
The number of integral points (integral point means both the coordinates should be integer exactly in the interior of the triangle with vertices \((0,0),(0,21)\) and \((21,0)\) is
1 133
2 190
3 233
4 105
Explanation:
(B) : Let the vertices of triangle be \(\mathrm{A}(21,0), \mathrm{B}(0,21)\) and \(\mathrm{O}(0,0)\) Thus, any point in the interior of the triangle lies in first quadrant. \(\therefore \quad a>0 \& b>0\) Point \((a, b)\) lies on the same side of the \(A B\) where lies. For \((0,0)\) therefore \(x+y \leq 21\) \(a+b-21\lt 0\) \(\Rightarrow \quad \mathrm{a}+\mathrm{b}\lt 21\) For \(\quad \mathrm{a}=1 ; \mathrm{b}\lt 21-1 \Rightarrow \mathrm{b}\lt 20\) \(b \in[1,19]\) total 19 integral value. For \(\quad a=2 ; b\lt 21-2 \Rightarrow b\lt 19\) Similarly \(b \in[1,18]\) total 18 integral values for \(\quad a=19 ; b\lt 21-19 \Rightarrow b\lt 2\) \(\mathrm{b}=1 \quad 1\) integral values Thus Number of integral points \(=19+18+\ldots .+1\) \(=\frac{19(19+1)}{2}=\frac{19 \times 20}{2}=190\) Thus there are total 190 integral points. Which lies inside the triangle.
VITEEE-2013
Co-Ordinate system
88281
In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio
1 \(2: 3: 5\)
2 \(1: 2: 3\)
3 \(1: 3: 7\)
4 \(3: 7: 9\)
Explanation:
(B) : We have, area of \(\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^{2}, \mathrm{~s}=\frac{3 \mathrm{a}}{2}\) In radius \(\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{a}}{2 \sqrt{3}}\) Circumradius \(\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^{3}}{\sqrt{3} \mathrm{a}^{2}}=\frac{\mathrm{a}}{\sqrt{3}}\) and exradius \(r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3} \mathrm{a}^{2} / 4}{\mathrm{a} / 2}\) \(=\frac{\sqrt{3}}{2} \mathrm{a}\) \(\therefore\) Required ratio \(=\mathrm{r}: \mathrm{R}: \mathrm{r}_{1}\) \(=\frac{\mathrm{a}}{2 \sqrt{3}}: \frac{\mathrm{a}}{\sqrt{3}}: \frac{\sqrt{3}}{2} \mathrm{a}=1: 2: 3\).
VITEEE-2011
Co-Ordinate system
88282
Let \(O=(0,0), A=(a, 11)\) and \(B=(b, 37)\) are the vertices of an equilateral triangle \(O A B\), then and batisfy the relation
1 \(\left(a^{2}+b^{2}\right)-4 a b=138\)
2 \(\left(a^{2}+b^{2}\right)-a b=124\)
3 \(\left(a^{2}+b^{2}\right)+3 a b=130\)
4 \(\left(a^{2}+b^{2}\right)-3 a b=138\)
Explanation:
(A) : Given, co-ordinate of vertices of equilateral triangle \(\mathrm{O}=(0,0), \mathrm{A}=(\mathrm{a}, 11), \mathrm{B}=(\mathrm{b}, 37)\) Let \(\mathrm{c}\) is the middle point of side \(\mathrm{AB}\) \(\therefore \quad \mathrm{C}=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, 24\right)\) \(\therefore \quad O C=\sqrt{3} C A\) \((\mathrm{OC})^{2}=(\sqrt{3}(\mathrm{CA}))^{2} \quad \text { [squaring both side] }\) \((\mathrm{OC})^{2}=3(\mathrm{CA})^{2}\) \(\left(\frac{\mathrm{a}+\mathrm{b}}{2}-0\right)^{2}+(24-0)^{2}=3\left[\left(\mathrm{a}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)^{2}+(11-24)^{2}\right]\) \(\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{4}+576=3\left[\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)^{2}+(-13)^{2}\right]\) \(\mathrm{a}^{2}+\mathrm{b}^{2}-4 \mathrm{ab}=138\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88279
If the vertices of a triangle are \(A(0,4,1), B(2,3\), \(-1)\) and \(C(4,5,0)\), then the orthocentre of \(\triangle \mathrm{ABC}\), is
1 \((4,5,0)\)
2 \((2,3,-1)\)
3 \((-2,3,-1)\)
4 \((2,0,2)\)
Explanation:
(B) : Vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(0,4,1), \mathrm{B}(2,3,-1)\) and \(\mathrm{C}(4,5,0)\). \(\mathrm{AB}=\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}}\) \(\mathrm{AB}=\sqrt{4+1+4}\) \(\mathrm{AB}=\sqrt{9}\) \(\mathrm{AB}=3\) \(\mathrm{BC}=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}}\) \(\mathrm{BC}=\sqrt{4+4+1}\) \(\mathrm{BC}=\sqrt{9}\) \(\mathrm{BC}=3\) \(\mathrm{CA}=\sqrt{(4-0)^{2}+(5-4)^{2}+(0-1)^{2}}\) \(\mathrm{CA}=\sqrt{16+1+1}\) \(\mathrm{CA}=3 \sqrt{2}\) \(\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}\) \(\therefore \triangle \mathrm{ABC}\) is a right angled triangle. We know that, the orthocenter of a right angled triangle is the vertex containing the right angle. \(\therefore\) Orthocenter is point B \((2,3,-1)\).
VITEEE-2014
Co-Ordinate system
88280
The number of integral points (integral point means both the coordinates should be integer exactly in the interior of the triangle with vertices \((0,0),(0,21)\) and \((21,0)\) is
1 133
2 190
3 233
4 105
Explanation:
(B) : Let the vertices of triangle be \(\mathrm{A}(21,0), \mathrm{B}(0,21)\) and \(\mathrm{O}(0,0)\) Thus, any point in the interior of the triangle lies in first quadrant. \(\therefore \quad a>0 \& b>0\) Point \((a, b)\) lies on the same side of the \(A B\) where lies. For \((0,0)\) therefore \(x+y \leq 21\) \(a+b-21\lt 0\) \(\Rightarrow \quad \mathrm{a}+\mathrm{b}\lt 21\) For \(\quad \mathrm{a}=1 ; \mathrm{b}\lt 21-1 \Rightarrow \mathrm{b}\lt 20\) \(b \in[1,19]\) total 19 integral value. For \(\quad a=2 ; b\lt 21-2 \Rightarrow b\lt 19\) Similarly \(b \in[1,18]\) total 18 integral values for \(\quad a=19 ; b\lt 21-19 \Rightarrow b\lt 2\) \(\mathrm{b}=1 \quad 1\) integral values Thus Number of integral points \(=19+18+\ldots .+1\) \(=\frac{19(19+1)}{2}=\frac{19 \times 20}{2}=190\) Thus there are total 190 integral points. Which lies inside the triangle.
VITEEE-2013
Co-Ordinate system
88281
In an equilateral triangle, the inradius, circumradius and one of the exradii are in the ratio
1 \(2: 3: 5\)
2 \(1: 2: 3\)
3 \(1: 3: 7\)
4 \(3: 7: 9\)
Explanation:
(B) : We have, area of \(\Delta=\frac{\sqrt{3}}{4} \mathrm{a}^{2}, \mathrm{~s}=\frac{3 \mathrm{a}}{2}\) In radius \(\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{\mathrm{a}}{2 \sqrt{3}}\) Circumradius \(\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}=\frac{\mathrm{a}^{3}}{\sqrt{3} \mathrm{a}^{2}}=\frac{\mathrm{a}}{\sqrt{3}}\) and exradius \(r_{1}=\frac{\Delta}{s-a}=\frac{\sqrt{3} \mathrm{a}^{2} / 4}{\mathrm{a} / 2}\) \(=\frac{\sqrt{3}}{2} \mathrm{a}\) \(\therefore\) Required ratio \(=\mathrm{r}: \mathrm{R}: \mathrm{r}_{1}\) \(=\frac{\mathrm{a}}{2 \sqrt{3}}: \frac{\mathrm{a}}{\sqrt{3}}: \frac{\sqrt{3}}{2} \mathrm{a}=1: 2: 3\).
VITEEE-2011
Co-Ordinate system
88282
Let \(O=(0,0), A=(a, 11)\) and \(B=(b, 37)\) are the vertices of an equilateral triangle \(O A B\), then and batisfy the relation
1 \(\left(a^{2}+b^{2}\right)-4 a b=138\)
2 \(\left(a^{2}+b^{2}\right)-a b=124\)
3 \(\left(a^{2}+b^{2}\right)+3 a b=130\)
4 \(\left(a^{2}+b^{2}\right)-3 a b=138\)
Explanation:
(A) : Given, co-ordinate of vertices of equilateral triangle \(\mathrm{O}=(0,0), \mathrm{A}=(\mathrm{a}, 11), \mathrm{B}=(\mathrm{b}, 37)\) Let \(\mathrm{c}\) is the middle point of side \(\mathrm{AB}\) \(\therefore \quad \mathrm{C}=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, 24\right)\) \(\therefore \quad O C=\sqrt{3} C A\) \((\mathrm{OC})^{2}=(\sqrt{3}(\mathrm{CA}))^{2} \quad \text { [squaring both side] }\) \((\mathrm{OC})^{2}=3(\mathrm{CA})^{2}\) \(\left(\frac{\mathrm{a}+\mathrm{b}}{2}-0\right)^{2}+(24-0)^{2}=3\left[\left(\mathrm{a}-\frac{\mathrm{a}+\mathrm{b}}{2}\right)^{2}+(11-24)^{2}\right]\) \(\frac{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab}}{4}+576=3\left[\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)^{2}+(-13)^{2}\right]\) \(\mathrm{a}^{2}+\mathrm{b}^{2}-4 \mathrm{ab}=138\)
