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enter image description here In the figure AD=BD B(-3,0),D(0,3) are given find Coordinates of point C I am new to analytic geometry I have tried to solve this and I found coordinates of point A
I think for solving this question we may need to use formula or theorems which I haven't studied yet that's why this question seems hard for me can you please help?

Clair
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  • For geometry problems like this there is a mantra. Draw extra lines! Draw extra lines! Here, draw those lines from $D$, parallel to the $x$-axis, until reaching $AC$, then downwards to the $y$-axis. – Parcly Taxel Feb 13 '18 at 19:29
  • @ParclyTaxel I think you meant to write downwards to the $x-$axis – Bergson Feb 13 '18 at 19:36
  • @ThomasBladt Yes, yes. – Parcly Taxel Feb 13 '18 at 19:39
  • I think the point of the question is for you to figure out the "formulas" you need. There are many ways to solve it; if you can find $A$ and know a little ordinary plane geometry you should already have the necessary tools. – David K Feb 13 '18 at 20:34

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COMMENT

►Let $A=(x,y)$ You have $(x,y)=(3,6)$ (Why?)

►Slope of ligne $\overline{BA}=1$ therefore slope of ligne $\overline{AC}=-1$ (Why?).

►ligne $AC: y=-x+9$ (Why?).

►Point $C$ has coordinates $(9,0)$ (Why?)

Piquito
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Note that D is the midpoint of AB thus

$$D=\frac{A+B}{2}\implies A=2D-B=2(0,3)-(-3,0)=(3,6)$$

Since $\angle B=45°$ C is symmetric to B with respect to A thus

$$x_A=\frac{x_B+x_C}{2}\implies x_C=2x_A-x_B=6-(-3)=9$$

therefore

$$C=(9,0)$$

user
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we get $$\vec{BD}=[3;3]$$ then $$\vec{BA}=[x_A+3;y_A-0]=2\vec{BD}$$ from here you will get $A(x_A,y_A)$

Dr. Sonnhard Graubner
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You say you have already found the coordinates of $A.$

There are formulas you could use such as the formula for the slope of a line perpendicular to a known line and the formula of a line of known slope through a known point, but you do not really need even those formulas for this question.

Try drawing a line through $A$ parallel to the $y$ axis. Let $E$ be the intersection of that line with the $x$ axis.

You now should be able to identify at least four right triangles in your diagram (though you do not really need that many). You can use similar triangles or any other tool you find convenient to find the lengths of edges of triangles parallel to the $x$ or $y$ axes. Eventually you want to find enough lengths so that you can determine the coordinates of $D$.

David K
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Hint: If to go from D to A you move $x$ across and $y$ up, then to move from A to C which is perpendicular you move by $\lambda y$ across and $-\lambda x$ up (in this case down, hence the negative). Find the proportionality ratio $\lambda$ that makes you meet the x-axis.

John Alexiou
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  • I am new to analytic I can't understand what are you saying – Clair Feb 14 '18 at 06:37
  • What I am saying is that the perpendicular to a vector ${\boldsymbol v}=\pmatrix{x,y}$ is ${\boldsymbol v}\perp=\pmatrix{y,-x}$. That's how you get to describe the leg _AC. – John Alexiou Feb 14 '18 at 14:07