# 10th Class Mathematics Coordinate Geometry Straight Lines

Straight Lines

Category : 10th Class

### Straight Lines

The locus of the points taken in one direction is called straight lines. The slope of the straight lines is the angle which the line make with the positive direction of x- axis. It is denoted by 'm' or' tan $\theta$'or' $\frac{dy}{dx}'$

Thus the most general form of the equation of the straight line having slope 'm' and y-intercept 'c' is given by:

Y = mx + c  Equation of the Line Passing Through One Point

Let $P({{X}_{1}},{{Y}_{1}})$ be the given point having slope 'm', then the equation of the line is given by $(Y-{{Y}_{1}})=m(X0-{{X}_{1}})$  Equation of Line Passing Through Two Given Points

Let $P({{X}_{1}},{{Y}_{2}})\,and\,Q({{X}_{2}},{{Y}_{2}})$ be the two points, then the equation of the line passing through P and Q is given by:

$(y-{{y}_{1}})=\left( \frac{{{Y}_{2}}-{{Y}_{1}}}{{{X}_{2}}-{{X}_{1}}} \right)\left( X-{{X}_{1}} \right)$  Intercepted Form of the Line

LET 'a' and 'b' be the x and y intercept of the line respectively, then the equation of line is given by:

$\frac{x}{a}+\frac{y}{b}=1$  Angle Between the Two Lines

If be the slope of the line and ' $\theta$' be the angle between the lines then the angle is given by:

$Tan\theta =\left[ \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right]$ If the lines are perpendicular, then ${{m}_{1}}{{m}_{2}}=-1$ and if the lines are parallel then ${{m}_{1}}{{m}_{2}}$ Distance of a Line From a Point

Let Ax + By + Cz = 0 be the equation of the line and 'd' is the distance of the line from the point $P({{X}_{1}},{{Y}_{1}})$, then distance of the line form the point is given by:

$d=\left| \left. \frac{A{{X}_{1}}+B{{X}_{1}}+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| \right.$  Distance Between Two Parallel Lines

The distance between the two parallel lines whose equation is AX + BY + ${{C}_{1}}$= 0 and AX + BY + ${{C}_{2}}$= 0 is given by:

$d=\left| \left. \frac{{{C}_{1}}+{{C}_{2}}}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| \right.$   Find the equation of the line passing through the points A (3, -2) and B (-1, 4).

(a) $3x+2y=-13$

(b) $3x-2y=13$

(c) $3x-2y=-5$

(d) $3x-2y=5$

(e) None of these

Explanation

The general equation of the straight line passing through the two points is given by:

$(y-{{y}_{1}})=\left( \frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)$

$(y+2)=\left( \frac{4+2}{3+2} \right)\left( x-3 \right)$

$\Rightarrow \,\,3x-2y=13$

Therefore, option (b) is correct and rest of the options is incorrect. The three vertices of a parallelogram ABCD, taken in order are A (1, -2), B (3, 6), & C (5, 10) find the coordinates of the fourth vertex D.

(a) (3, 2)

(b) (3, 2)

(c) (3, 4)

(d) (5, -9)

(e) None of these

Let A (1, -2) B (3, 6) C (5, 10) be the given vertices of the parallelogram ABCD & let D (a, b) be its fourth vertex join AC & BD.

Let AC & BD intersect at the point 0.

We know that the diagonals of a parallelogram bisect each so, O is the mid-point of AC as well as that of BD.

Mid-point of AC is $\left( \frac{1+5}{2},\frac{-2+10}{2} \right)$ i.e. (3, 4).

Mid-point of BD is $\left( \frac{a+3}{2},\frac{b+6}{2} \right)$

$\therefore \,\,\,\frac{3+a}{2}=3\And \frac{6+b}{2}=4$

$\Rightarrow \,\,a=3\And b=2$ If the angle between the two lines is $\frac{\pi }{4}$ and the slope of one line is 3, then the slope of the other line is given by:

(a) $\left( 2,\frac{1}{2} \right)$

(b) $\left( \frac{2}{3},\frac{1}{2} \right)$

(c) $\left( \frac{2}{3},1 \right)$

(d) $\left( -2,\frac{1}{2} \right)$

(e) None of these

Explanation

Let the slope of the second line be 'm'

Then the angle between the line is given by:

$Tan\theta =\left[ \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right]$

$\Rightarrow \,\,Tan\frac{\pi }{4}=\left[ \frac{m-3}{1+3m} \right]$

$\Rightarrow \,\,\,m=-2\,,\,and\frac{1}{2}$ Find the equation of the line which is perpendicular to the line segment joining the points A (1, 0) and B (2, 3) and divide it in the ratio 1 : k.

(a) $\left( Y-\frac{3}{1+k} \right)=\frac{-1}{3}\left( X-\frac{2+k}{1+k} \right)$

(b) $\left( Y+\frac{3}{1+k} \right)=\frac{1}{3}\left( X-\frac{2+k}{1+k} \right)$

(c) $\left( Y+\frac{3}{1+k} \right)=\frac{-1}{3}\left( X+\frac{2+k}{1+k} \right)$

(d) $\left( Y+\frac{3}{1+k} \right)=4\left( X+\frac{2+k}{1+k} \right)$

(e) None of these

Explanation

We have the slope of the line $AB=\frac{3-0}{2-1}=3$

Therefore, slope of the line perpendicular to $AB=\frac{-1}{3}$

The coordinate of the point which divide the line segment joining AB in the ratio 1: k is given by:

$X=\frac{1\times 2+k\times 1}{1+k}\,\,and\,y=\frac{0\times k+3\times 1}{1+k}$

$\Rightarrow \,\,\,X=\frac{2+k}{1+k}\,and\,\,Y=\frac{3}{1+k}$

Hence the equation of the line perpendicular to the line AB is given by:

$\left( Y-\frac{3}{1+k} \right)=\frac{-1}{3}\left( X-\frac{2+k}{1+k} \right)$ The owner of a diary farm finds that he can sell 980 litres of milk each month at the rate of Rs. 14 per litres and 1220 litres per month at the rate of Rs 16 per litres. Assuming that he can sell all the milk he stores, find the quantity of milk he should store to sell it at the rate of Rs. 17 per litres per month.

(a) 1340 litres

(b) 1560 litres

(c) 1785 litres

(d) 1985 litres

(e) None of these Find the angle between the lines whose equations are 5x + 3y = 4 and 4x + 5y = 6.

(a) 1

(b) $\frac{1}{\sqrt{3}}$

(c) $Ta{{n}^{-1}}\left( \frac{13}{21} \right)$

(d) $\sqrt{3}$

(e) None of these • The Greek mathematician Menaechmus solved problems and proved theorems by using a method that had a strong resemblance to the use of coordinates.
• The eleventh century Persian mathematician Omar Khayyam saw a strong relationship between geometry and algebra.
• Dilation - a transformation that occurs from a center point.
• Rotation - a transformation that is the composite of two reflections with respect to two intersecting lines.
• Oblique - not at right angles. • In a three dimensional system there are three axes.
• The three coordinate planes divide the space into the eight parts called octants.
• The distance between the two points is given by: $AB=\sqrt{{{({{x}^{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}$
• he section-formula for the line joining two points in the ratio m:n is given by: $X=\frac{m{{X}_{1}}+n{{X}_{2}}}{m+n}\,and\,Y=\frac{m{{Y}_{1}}+n{{Y}_{2}}}{m+n}$
• The most general equation of the straight line is given by Ax + By + C = 0.

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