Straight Lines and Lines Joining Origin to Intersection Points - Class XI

Solving the geometry: The line through (1,2) at angle θ to +x-axis meets x+y=4 at distance √(2/3). Using parametric form and distance formula gives θ = 15° as the valid solution. This involves setting up the line equation, finding intersection point, and applying distance constraint.

Let the given point  be $A=(−5,4)$ and the given lines be $l _1 \rightarrow x+y+1=0$ and  $l _2\rightarrow x+y-1=0$

The  point $A$ lies on $l _1$

If segment $AM\perp l _2$ and $M$ lies on $l _2$, then, the distance $ AM$ is given by,

$\Rightarrow AM=\dfrac{|−5+4−1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt 2}=\sqrt 2$

$\Rightarrow $ This means that if $B$ is any point  on $l _2$ then  $AB>AM$. No line other than $AM$ cuts off an intercept of

length $\sqrt 2$ between $l _1$ and $l _2$.

$\Rightarrow $To determine the equation of $AM$, we need to find the co-ordinates of the Point $M$

Since, $AM\perp l _2$ and  the slope $l _2$ is $−1$, the slope of$AM$ must be $1$.  Also  $A(−5,4)$ lies on $AM$

By the point slope formula, the equation of the required line is

$\Rightarrow  y−4=1(x−(−5))$

$\Rightarrow y-4=x+5$

$\Rightarrow x−y+9=0$

The slopes of the given lines are m₁ = 3/4 and m₂ = 12/5. For a line to be equally inclined to both, its slope m must satisfy |(m-m₁)/(1+mm₁)| = |(m-m₂)/(1+mm₂)|. Solving gives two possible slopes: m = -7/9 (internal bisector) or m = 9/7 (external bisector). Using point (4,5) with slope -7/9: y-5 = (-7/9)(x-4), giving 9y-45 = -7x+28, or 7x+9y=73.

Substituting y = 4x + c into x² + 4y² = 4 gives: x² + 4(4x+c)² = 4, or x² + 4(16x² + 8cx + c²) = 4. This simplifies to 65x² + 32cx + 4c² - 4 = 0. For the line to be tangent to the ellipse, this quadratic must have exactly one solution, so discriminant = 0: (32c)² - 4(65)(4c²-4) = 0. This gives 1024c² - 1040c² + 1040 = 0, or -16c² + 1040 = 0, giving c² = 65. Therefore c = ±√65, so there are 2 values.

Let the given point be $A=(-5,4)$ and the given lines be,

Let the given point be $A\left( {{x} _{1}},{{y} _{1}} \right)=\left( -5,4 \right)$ and the given lines be

$ x+y+1=0\,\,......\,\,\left( 1 \right) $

$ x+y-1=0\,\,......\,\,\left( 2 \right) $

Observe that,

From equation (1) to,

Let AM is perpendicular distance

Then, $AM=\dfrac{\left| -5+4-1 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

Let B is any point on equation $(2)$to,

From equation (2)

$ x+y-1=0 $

$ y=-x+1 $

On comparing that,

$y=mx+c$

Then, $m=-1$

Slope of perpendicular line is

${{m} _{1}}=\dfrac{-1}{m}=1$

Then, equation of line is

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

$ \Rightarrow y-4=1\left( x+5 \right) $

$ \Rightarrow y-4=x+5 $

$ \Rightarrow x-y+5+4=0 $

$ \Rightarrow x-y+9=0 $

Hence, this is the
answer.

Solving x+y=4 and x-y=2 gives the intersection point (3,1). A line through this point with slope 3/4 has equation: y-1 = (3/4)(x-3), or 4y-4 = 3x-9, or 3x-4y=5. Substituting into y²=4(x-3) gives a quadratic. The x-coordinates of the intersection points are found by solving for the parameter along the line. The difference |x₁ - x₂| = 16/9.

As the line $x+y=1$ intersect the curve,$\displaystyle x^{2}+y^{2}+x-2y-m=0$

Joint equation of the lines joining the origin and the point of intersection of the line $y =mx + 2$ and
the curve $x^{2} + y^{2}=1$ is
$ \displaystyle x^{2} +y^{2} =\left( \frac{y-mx}{2}\right)^{2} $
$ x^{2}(4 -m^{2} )+2mxy +3y^{2} = 0$
Since these lines are at right angles
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

Homogenizing the curve with the help of the straight line.

By application of the method of homogenization we get
$3x^2+mxy-4x(2x+y)+1(2x+y)^{2}$
$=3x^2+mxy-8x^2-4xy+4x^2+y^2+4xy$
$=-x^2+mxy+y^2$
$=0$
Hence the equations of the lines is given by $ x^2-mxy-y^2=0$
Hence the lines are perpendicular for all values of $m.$

Given equations of curve and the line are $2x^2 + 3xy -4x + 1 = 0$ and $3x + y = 1$.
Homogenising the curve with the line gives
$2x^2+3xy-4x(3x+y)+(3x+y)^2=0$
$\Rightarrow 2x^2+3xy-12x^2-4xy+9x^2+y^2+6xy=0$
$\Rightarrow y^2-x^2+5xy=0$
$\therefore$ The equation of the lines joining the origin to the points of intersection of the curve and hte line is $x^2-5xy-

y^2=0$
Hence, option A.