Question

1. show that points(2,-3), (5,0), (2,3), and (-1,0), are the vertices of a square.
2. the distance (x,1) is 2 square root of 5 units from (2, 3), find x.
3. what are the coordinates of the point 3 units from the Y-axis and at distance square root 5 from (5,3)?
4. what is the equation of a line through (7,-3) and perpendicular to the line whose inclination is Arctan 2/3.
5. show that lines 2x + 3Y - 2 =0, 3X - 2Y + 23 =0 and X - 5Y = 12 = 0 are the sides of an isosceles triangle.

Answer 1

Answer on Question #65433 – Math | Geometry

1. show that points A(2,-3), B(5,0), C(2,3), and D(-1,0), are the vertices of a square.

Lets calculate length of diagonals BD and AC; $AC = \sqrt{(2 - 2)^2 + (3 + 3)^2} = 6$;

$BD = \sqrt{(5 + 1)^2 + 0} = 6$ As, two diagonals are equal: quadrangle is square;

2. the distance $(x,1)$ is 2 square root of 5 units from $(2,3)$ , find $x$

$$\sqrt {(x - 2) ^ {2} + (1 - 3) ^ {2}} = 2 \sqrt {5};$$ $$(x - 2) ^ {2} = 16;$$ $$x - 2 = \pm 4;$$ $$x = - 2, x = 6$$

3. what are the coordinates of the point 3 units from the Y-axis and at distance square root 5 from (5,3)?

As the point is 3 units from Y-axis: its x component equals 3 or -3;

$$\sqrt {(\pm 3 - 5) ^ {2} + (y - 3) ^ {2}} = \sqrt {5}$$ $$8 ^ {2} + (y - 3) ^ {2} = 5 \quad 2 ^ {2} + (y - 3) ^ {2} = 5$$ $$y \in \emptyset \quad (y - 3) ^ {2} = 1$$ $$y - 3 = \pm 1$$ $$y = 4, y = 2$$

Answer: (3,4) or (3,2)

4. what is the equation of a line through A(7,-3) and perpendicular to the line whose inclination is Arctan 2/3.

Let $y = k_{1}x + b_{1}$ — equation of first line

inclination: $\arctan 2 / 3 \Rightarrow k_1 = \frac{2}{3}$ ;

Let $y = k_{2}x + b_{2}$ — equation of second line, if line1 is perpendicular to line2; $k_{2} = -\frac{1}{k_{1}} = -\frac{3}{2}$ ;

$$y = - \frac {3}{2} x + b _ {2}; \mathrm {A} \in \text {line2}; \quad \mathrm {y} _ {\mathrm {A}} = - \frac {3}{2} x _ {\mathrm {A}} + b _ {2};$$ $$- 3 = - \frac {3}{2} * 7 + b _ {2};$$ $$b _ {2} = \frac {1 5}{2};$$

So, equation of line2: $\mathbf{y} = -\frac{3}{2}\mathbf{x} + \frac{15}{2}$

5. show that lines $2x + 3Y - 2 = 0$, $3X - 2Y + 23 = 0$ and $X - 5Y = 12 = 0$ are the sides of an isosceles triangle.

Let vector $\vec{a}, \vec{b}, \vec{c}$, parallel to line 1,2,3; $\vec{a} = (2,3)$; $\vec{b} = (3, -2)$; $\vec{c} = (1, -5)$

Lets find angles between vector $\vec{a}, \vec{c}$ and $\vec{b}, \vec{c}$;

$$\varphi (\vec {a}, \vec {c}) = \frac {\vec {a} \vec {c}}{| \vec {a} | | \vec {c} |} = \frac {2 * 1 - 5 * 3}{13 \sqrt {2}} = - \frac {1}{\sqrt {2}}; \varphi = 135 {}^ {\circ};$$ $$\varphi (\vec {b}, \vec {c}) = \frac {\vec {b} \vec {c}}{| \vec {b} | | \vec {c} |} = \frac {3 * 1 + 2 * 5}{13 \sqrt {2}} = \frac {1}{\sqrt {2}}; \varphi = 45 {}^ {\circ};$$

Angle between vectors $\mathbf{a}$ and $\mathbf{c} = 135{}^{\circ}(45{}^{\circ})$

Angle between vectors $\mathbf{b}$ and $\mathbf{c} = 45{}^{\circ}$

As angles between vectors are same, angles between lines are same, so the triangle is isosceles;

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