>> Elementary Mathematics >> Straight lines

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1). The distance of the point (-2, 3) from the line x -y = 5 is:
The distance of the point \( \Large \left(-2,\ 3\right) \) from the line \( \Large x-y=5 \) is | ||||

2). The equation of the straight line joining the origin to the point of intersection of \( \Large y-x+7=0\ and\ y+2x-2=0 \)is:
The intersection point of \( \Large y-x+7=0 \) and \( \Large y+2x-2=0 \) is \( \Large \left(3,\ -4\right) \) Therefore, Equation of straight line joining from origin to the point \( \Large \left(3,\ -4\right) \) is \( \Large y-0 = \frac{-4}{3} \left(x-0\right) \) => \( \Large 3y = -4x => 4x+3y=0 \) | ||||

3). The equation of the straight line which is perpendicular to\( \Large y = x \) and passes through (3, 2) is:
Because, Slope of given line y = x is 1 Therefore, Slope of required line which is perpendicular to given line is -1 Thus, the equation of required 1 line passing through \( \Large \left(3,\ 2\right) \) and slop -1 is \( \Large y-2 = -1 \left(x-3\right) => x+y=5 \) | ||||

4). The angle between the straight lines \( \Large x-y\sqrt{3}=5\ and\ \sqrt{3}x+y=7 \) is:
Given equation is compared with \( \Large a_{1}x+b_{1}y=0\ and\ a_{2}x+b_{2}y=0 \) | ||||

5). The three straight lines \( \Large ax+by=c,\ bx+cy=a\ and\ cx+ay=b \) are collinear if:
We have \( \Large ax+by=c \) ...(i) \( \Large bx+cy=a \) ...(ii) and \( \Large cx+ay=b \) ...(iii) On adding equation (i), (ii) and (iii), we get \( \Large ax+by+bx+cy+cx+ay = a+b+c \) => \( \Large \left(a+b+c\right)x+ \left(a+b+c\right)y = \left(a+b+c\right) \) On comparing with \( \Large ox+oy=0 \) (for collinearity) We get\( \Large a+b+c = 0 \) | ||||

6). If a tangent to the curve \( \Large y=6x-x^{2} \) is parallel to the line \( \Large 4x-2y-1=0 \) then the point of tangency on the curve is: ' _
Since, tangent is parallel to \( \Large y = 2x-\frac{1}{2} \) | ||||

7). Distance between the lines \( \Large 5x+3y-7=0\ and\ 15x+9y+14=0 \) is:
Given equation of lines are | ||||

8). The equation of the sides of a triangle are \( \Large x-3y=0.\ 4x+3y=5\ and\ 3x+y=0 \). The lines \( \Large 3x-4y=0 \) passage through
Two sides \( \Large x-3y=0\ and\ 3x+y=0 \) are perpendicular to each other. Therefore its orthocentre is the point of intersection of \( \Large x-3y=0 \) i.e. \( \Large \left(0,\ 0\right) \) so, the line \( \Large 3x-4y=0 \) passes through the orthocentre of triangle. | ||||

9). If (-4, -5) is one vertex and \( \Large 7x-y+8=0 \) is one diagonal of a square then the equation of second diagonal is:
Equation of perpendicular line to \( \Large 7x-y+8=0 \) is \( \Large x+7y=h \) which is passes through \( \Large \left(-4,\ 5\right) \) h = 31 So, equation of another diagonal is \( \Large x+7y=31 \) | ||||

10). The number of the straight which is equally inclined to both the axes is
There are four possible straight line which are equally inclined in both the axes. i.e., Ist, IInd, IIIrd and IVth quadrant. |