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Question Number 99877 by I want to learn more last updated on 23/Jun/20

Commented by bobhans last updated on 24/Jun/20

$$(3.1)\mathrm{gradient}\mathrm{of}\mathrm{PQ}=\tan {45}^{\mathrm{o}}=1$$$$(3.2)\mathrm{MN}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathrm{PQ}\Rightarrow \mathrm{then}\mathrm{equation}$$$$\mathrm{of}\mathrm{MN}\Rightarrow \mathrm{y}=1.(\mathrm{x}-7)+1;\mathrm{y}=\mathrm{x}-6$$$$(3.3)\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{MN}=\frac{1}{2}\mathrm{QP}=\frac{7\sqrt{2}}{2}$$$$(3.4)\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{RS}=7\sqrt{2}$$$$(3.6)\mathrm{coordinates}\mathrm{of}\mathrm{P}\Rightarrow \frac{\mathrm{b}+3}{\mathrm{a}+2}=1;\mathrm{b}+3=\mathrm{a}+2$$$$\Rightarrow \sqrt{{(\mathrm{b}+3)}^2+{(\mathrm{a}+2)}^2}=7\sqrt{2};2{(\mathrm{a}+2)}^2=49.2$$$$\Rightarrow \{\begin{array}{l}\mathrm{a}=5\\ \mathrm{b}=4\end{array}\Rightarrow \mathrm{P}(5,4)$$$$$$

Commented by bobhans last updated on 24/Jun/20

$$\mathrm{coordinates}\mathrm{of}\mathrm{R}(\mathrm{x},\mathrm{y})\Rightarrow \{\begin{array}{l}7=\frac{\mathrm{x}+\mathrm{a}}{2}\\ 1=\frac{\mathrm{y}+\mathrm{b}}{2}\end{array}$$$$\mathrm{x}=14-\mathrm{a}=14-5=9$$$$\mathrm{y}=2-\mathrm{b}=2-4=-2.\mathrm{coordinates}\mathrm{of}\mathrm{R}(9,-2)$$

Commented by I want to learn more last updated on 24/Jun/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.\mathrm{But}(3.5)\mathrm{is}\mathrm{misssing}.\mathrm{please}\mathrm{help}$$

Commented by bobhans last updated on 24/Jun/20

$$\mathrm{gradient}\mathrm{of}\mathrm{RS}=1.\mathrm{let}\mathrm{point}\mathrm{S}(\mathrm{x},\mathrm{y})$$$$\frac{\mathrm{y}+2}{\mathrm{x}-9}=1,\mathrm{y}+2=\mathrm{x}-9$$$$\mathrm{length}\mathrm{of}\mathrm{RS}=7\sqrt{2}\Rightarrow \sqrt{{(\mathrm{y}+2)}^2+{(\mathrm{x}-9)}^2}=7\sqrt{2}$$$$2{(\mathrm{x}-9)}^2=49.2\Rightarrow \mathrm{x}=16\wedge \mathrm{y}=5$$$$\mathrm{we}\mathrm{get}\mathrm{coordinates}\mathrm{of}\mathrm{S}(16,5)$$

Commented by I want to learn more last updated on 24/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

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