First Quadrant Triangle

Let $m$ be the gradient of the straight line, with $m<0$ , then the equation of the straight line is $\frac {y-3}{x-4} = m$ or $y = mx + 3 - 4m$

At $x$ -intercept, $y = 0$ , $x$ -intercept is $\frac {4m-3}{m}$

At $y$ -intercept, $x = 0$ , $x$ -intercept is $3 - 4m$

Area of triangle equals to $\frac {1}{2} \times ( x- \text{intercept} )\times ( y- \text{intercept} ) = \frac {1}{2} \cdot \frac {4m-3}{m} \cdot (3-4m) = - \frac {1}{2} \cdot \frac {(4m-3)^2} {m}$

To minimize the area of triangle, we find the derivative of the function $f(m) = - \frac {1}{2} \cdot \frac {(4m-3)^2} {m}$

$\Rightarrow f(m) = -\frac{1}{2} (16m - 24 + \frac {9}{m} ) \Rightarrow f'(m) = -\frac{1}{2} (16 - \frac {9}{m^2} )$ , when $f'(m) = 0$ , $m = \pm \frac {3}{4}$ , but because $m < 0$ , $m = - \frac {3}{4}$ only, and because $f''(- \frac {3}{4}) > 0$ , the area of triangle is minimized when $m = - \frac {3}{4}$ . Substitute in the value, we get the area of triangle as $\boxed{24}$

The equation of the line can be represented as $y - 3 = m(x - 4)$ , where $m < 0$ . Thus, our x and y intercepts are $-\frac{3}{m} + 4$ and $-4m + 3$ , respectively, making our area of the triangle

$2A = -\frac{1}{m} (-4m + 3)(-4m + 3)$

$2A = -16m + 24 - \frac{9}{m}$

Using calculus,

$\frac{dA}{dm} = -16 + \frac{9}{m^2}$ , so extrema are at $m = \pm \frac{3}{4}$ , and since $m < 0$ then $m = -\frac{3}{4}$ . We can use the second derivative test to confirm this as a minimum.

Thus, the minimum area is $A = \frac{1}{2} -16(-\frac{3}{4}) - \frac{9}{-\frac{3}{4}} + 24) = 24$

We know that the general form of the equation of a line is ${y}={mx}+{c}$

Let c be the height of the triangle and

Let the base be the x intercept i.e $x=\frac{-c}{m}$

The line passes through the point (4,3) and substituting this point into the general equation, ${3}={4m}+{c}$

thus ${m}=\frac{3-c}{4}$

Area of a triangle equals; ${Area}=\frac{1}{2}{bh}$

Thus $A_{triangle}=\frac{1}{2}\frac{-c}{m}\times{c}$

substituting for m into the above equation we get $A_{triangle}=\frac{2c^2}{c-m}$

Differentiating A with respect to c we get $\frac{dA}{dc}=\frac{2c^2-12c}{(c-m)^2}$

But we know that for a max. or min. value the derivative = 0 .

Thus $\frac{dA}{dc}=\frac{2c^2-12c}{(c-m)^2}={0}$

from which we get; ${2c^2}-{12c}={0}$

${c}={6}$

substituting for c into $A_{triangle}=\frac{2c^2}{c-m}$

i.e $A_{triangle}=\frac{2(6)^2}{(6)-m}$

$A_{triangle}={24}$

Therefore the minimum area of the triangle equals $\boxed{24sq. units}$

Let the slow of line be m Then eqn of line is y-4=m(x-3) Convert this into xy intercept form..... Then area is .5* x intercept* y intercept. .....

The expression is (4m-3)^2/2m

Using differenciation..... the min value is 24

The minimum area is obtained only when the optimum angle between any line -that passes through a certain point- and X axis should equal 45 degree .. In the case of first quarter the angle equals 135 degree .. now you have the Slope ( = tan(135) = -1) and a point (4,3) so the equation of this line is: (Y-Y1)= Slope * (X-X1) (Y-3)= - (X-4) assume X= 0 and get Y Y-3=4 Y= 7 (0,7) then Y=0 and get X 0-3 = -X+4 X=7 (7,0) so you have the intersection point with the two axis the area of this triangle = 0.577 = 24.5 ...

the solution is 24

The minimum area is obtained only when the optimum angle between any line -that passes through a certain point- and X axis should equal 45 degree .. In the case of first quarter the angle equals 135 degree .. now you have the Slope ( = tan(135) = -1) and a point (4,3) so the equation of this line is: (Y-Y1)= Slope * (X-X1) (Y-3)= - (X-4) assume X= 0 and get Y Y-3=4 Y= 7 (0,7) then Y=0 and get X 0-3 = -X+4 X=7 (7,0) so you have the intersection point with the two axis the area of this triangle = 0.5 7 7 = 24.5 ...

the solution is 24

Call x,y are two sides of triangle and x is in horizontal axis. Use Thales theory, we have 4/x+3/y=1. And xy=3x+4y so we can infer that xy>=48 or xy/2>=24

Let the line be: x/a + y/b=1; hence 4/a + 3/b = 1 or b = 3a/(a-4). Tr. Area = (3/2)a^2/(a-4) which is minimum at x=8 or Min A = (3/2)*8^2/(8-4) = 24 sq. Units