Question

Show that ${x^2} - 6x + 5{y^2} + 10x - 14y + 9 = 0$. Find the acute angle?

Answer 1

The general equation of a straight line is $y = mx + c$, where m is the gradient, and y = c is the value where the line cuts the y-axis. This number c is called the intercept on the y-axis. The equation of a straight line with gradient m and intercept c on the y-axis is $y = mx + c$.

Complete step-by-step solution:
Given
$\therefore {x^2} - 6x + 5{y^2} + 10x - 14y + 9 = 0$
Comparing with
$\therefore a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
We get that
$a = 1,\,h = - 3,\,\,b = 5,\,g = 5,\,f = - 7,\,c = 9$
We know that
\therefore \left[ {\begin{array}{*{20}{c}} a&h;&g; \\\ h&b;&f; \\\ g&f;&c; \end{array}} \right]
.Put the value
= \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&5 \\\ { - 3}&5&{ - 7} \\\ 5&{ - 7}&9 \end{array}} \right]
Simplify
$= 1\\{ (9 \times 5) - ( - 7 \times - 7)\\} + 3\\{ (9 \times - 3) - (5 \times - 7)\\} + 5\\{ ( - 3 \times - 7) - (5 \times 5)\\}$
$= (45 - 49) + 3( - 27 + 35) + 5(21 - 25)$
$= - 4 + 3 \times 8 + 5 \times - 4$
$= 24 - 24$
$= 0$
Given equation present the equation of line
Now
As we know that
$\therefore \tan \theta = \left[ {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right]$
Put the value
$\Rightarrow \tan \theta = \left[ {\dfrac{{2\sqrt {{{( - 3)}^2} - (1 \times 5)} }}{{1 + 5}}} \right]$
$\Rightarrow \tan \theta = \left[ {\dfrac{{2\sqrt {9 - 5} }}{6}} \right]$
$\Rightarrow \tan \theta = \left[ {\dfrac{{\sqrt 4 }}{3}} \right]$
$\Rightarrow \tan \theta = \left( {\dfrac{2}{3}} \right)$
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right)$

Note: The slope of a line is a number that describes both the direction and the steepness of the line. Slope is often denoted by the letter ‘m’. The line is increasing. It goes up from left to right. The slope is positive. A line is decreasing if it goes down from left to right. The slope is negative. If a line is horizontal the slope is zero. This is a constant function.